Lesson 5

Trigonometric Functions, and Identities

Radians, arcs and sectors, the circular functions of any angle and their graphs, the sine and cosine rules, trigonometric equations, the standard identities, small angles and further properties of triangles.

Taught 14 Sept 2026

Measuring Angles

Measurement of Rotation

When a line $OP$ is pivoted at $O$ and rotates from its initial position $OP_0$ to a new position $OP_1$ , the angle $P_0OP_1$ is a measure of the rotation of $OP$ .

Figure 5.1. The angle

\theta

measures the rotation of

OP

from

OP_0

OP_1

The angle $\theta$ is usually measured in one of two units.

The Degree

The ancient Babylonian mathematicians, who thought that the solar year was $360$ days long, divided one complete revolution into $360$ equal parts, each part now being known as one degree, $1^\circ$ . Using the degree as the unit of rotation, half a revolution corresponds to $180^\circ$ and a quarter of a revolution, that is a right angle, corresponds to $90^\circ$ .

Angles smaller than a degree are usually given as decimal parts, so that half a degree is $0.5^\circ$ . In some fields, such as navigation, a degree is divided into $60$ minutes, written $60'$ , and each minute into $60$ seconds, written $60''$ .

The Radian

Definition 5.1 (Radian).

If an arc $PP_1$ of a circle with centre $O$ is equal in length to the radius of the circle, the angle $POP_1$ is one radian, written $1^{\mathrm{c}}$ .

Figure 5.2. An arc equal in length to the radius subtends one radian at the centre.

The number of radians in one complete revolution is therefore

\frac{\text{circumference}}{\text{radius}} = \frac{2\pi r}{r} = 2\pi,

since the circumference of a circle of radius $r$ is $2\pi r$ . So

2\pi \text{ radians} = 360^\circ, \qquad \pi \text{ radians} = 180^\circ, \qquad \frac{\pi}{2} \text{ radians} = 90^\circ .

When an angle is quoted in terms of $\pi$ it is normal to omit the radian symbol, so we write $180^\circ = \pi$ , not $\pi^{\mathrm{c}}$ .

Angles which are simple fractions of $180^\circ$ are easily expressed in radians in terms of $\pi$ , using $180^\circ = \pi$ , and conversely:

\begin{aligned} 60^\circ &= 60 \times \frac{\pi}{180} = \frac{\pi}{3}, & 135^\circ &= 135 \times \frac{\pi}{180} = \frac{3\pi}{4}, \\ \frac{7\pi}{6} &= \frac{7\pi}{6} \times \frac{180^\circ}{\pi} = 210^\circ, & \frac{5\pi}{3} &= \frac{5\pi}{3} \times \frac{180^\circ}{\pi} = 300^\circ . \end{aligned}

Converting in this way is not convenient for angles which are not simple fractions of a revolution. It would not be easy, for instance, to express $47^\circ 34'$ as a multiple of $\pi$ . We would first have to express $34'$ as a decimal part of a degree, $34' = 0.567^\circ$ , and then use

47.567^\circ = 47.567 \times \frac{\pi}{180} = 0.830 \text{ radians}.

Calculations like this are tedious, and a calculator which offers the conversion avoids them.

To visualise the size of an angle of one radian, it helps to remember that $\pi$ radians $= 180^\circ$ and that $\pi = 3.14$ to two decimal places, so

1 \text{ radian} = \frac{180^\circ}{3.14} = 57.3^\circ \quad\text{to three significant figures},

a little less than $60^\circ$ .

Problem 5.1.

Without a calculator, express $30^\circ$ , $270^\circ$ and $22.5^\circ$ in radians, in terms of $\pi$ .
Without a calculator, express $\dfrac{5\pi}{6}$ , $\dfrac{11\pi}{6}$ and $\dfrac{4\pi}{9}$ in degrees.
Express $54^\circ 45'$ in radians, and $2.86$ radians in degrees and minutes.

The Circular Functions

The Ratios of an Acute Angle

For any acute angle $\theta$ there are six trigonometric ratios, each of which is defined by referring to a right-angled triangle containing $\theta$ .

Figure 5.3. A right-angled triangle

OAB

containing the acute angle

\theta

O

Definition 5.2 (Trigonometric Ratios of an Acute Angle).

In a triangle $OAB$ with a right angle at $A$ and an acute angle $\theta$ at $O$ ,

\begin{aligned} \sin\theta &= \frac{AB}{OB}, &\quad \operatorname{cosec}\theta &= \frac{OB}{AB}, \\ \cos\theta &= \frac{OA}{OB}, &\quad \sec\theta &= \frac{OB}{OA}, \\ \tan\theta &= \frac{AB}{OA}, &\quad \cot\theta &= \frac{OA}{AB} . \end{aligned}

These are the sine, cosine, tangent, cosecant, secant and cotangent of $\theta$ .

Each ratio has a single value for any one acute angle, since all right-angled triangles containing $\theta$ are similar, and the values are available from a calculator.

Length of an Arc and Area of a Sector

The circumference of a circle of radius $r$ is $2\pi r$ and its area is $\pi r^2$ , and these formulae can be used to derive further results.

Consider an arc which subtends an angle $\theta$ at the centre of the circle, where $\theta$ is measured in radians. From the definition of a radian, the arc which subtends $1$ radian at the centre has length $r$ , so an arc which subtends $\theta$ radians at the centre has length $r\theta$ .

The area of a sector containing an angle of $\theta$ radians at the centre can be found by regarding the sector as a fraction of the circle. The ratio of the area of the sector to the area of the circle is equal to the ratio of the angle $\theta$ contained in the sector to the angle $2\pi$ contained in the whole circle, that is

\frac{\text{area of sector}}{\pi r^2} = \frac{\theta}{2\pi}, \qquad\text{so}\qquad \text{area of sector} = \tfrac12 r^2\theta .

Figure 5.4. A sector

AOB

containing the angle

\theta

at the centre

O

Hence if an arc $AB$ subtends an angle of $\theta$ radians at the centre $O$ of a circle of radius $r$ ,

\text{length of arc } AB = r\theta, \qquad \text{area of sector } AOB = \tfrac12 r^2\theta .

Example 5.3.

A railway line changes direction by $20^\circ$ when passing round a circular arc of length $500$ m. What is the radius of the arc?

The line turns through the angle which the arc subtends at the centre, and in radians $20^\circ = 20 \times \dfrac{\pi}{180} = \dfrac{\pi}{9}$ . Using length of arc $= r\theta$ , with $\theta$ in radians,

500 = r \times \frac{\pi}{9}, \qquad\text{so}\qquad r = \frac{4500}{\pi} = 1432 .

The radius of the track is therefore $1432$ m.

Example 5.4.

A chord $AB$ divides a circle of radius $2$ m into two segments. If $AB$ subtends an angle of $60^\circ$ at the centre $O$ of the circle, find the area of the minor segment.

The triangle $AOB$ has base $OA = 2$ and height $2\sin 60^\circ$ , so

\text{area of triangle } AOB = \tfrac12 \times 2 \times 2\sin 60^\circ = 1.732\ \text{m}^2 .

Since $60^\circ = \dfrac{\pi}{3}$ radians,

\text{area of sector } AOB = \tfrac12 \times 2^2 \times \frac{\pi}{3} = \frac{2\pi}{3} = 2.094\ \text{m}^2 .

So the area of the minor segment, shaded in Figure 5.5, is $(2.094 - 1.732)\ \text{m}^2 = 0.362\ \text{m}^2$ .

Figure 5.5. The minor segment cut off by a chord subtending

60^\circ

at the centre.

Example 5.5.

Two discs of radii $3$ cm and $4$ cm are laid on a table with their centres $5$ cm apart. Find the perimeter of the figure-eight shape so formed.

Let the discs have their centres at $A$ and $B$ and cross at $C$ . Since $5^2 = 3^2 + 4^2$ , the triangle $ABC$ is right-angled at $C$ . Let $\alpha$ be the angle $CAB$ and $\beta$ the angle $CBA$ . Then

\tan\alpha = \tfrac43, \quad \alpha = 53.13^\circ = 0.927 \text{ radians}, \qquad \beta = 90^\circ - \alpha = 36.87^\circ = 0.644 \text{ radians}.

The perimeter is made up of an arc subtending $2\pi - 2\alpha$ radians in the circle of radius $3$ cm and an arc subtending $2\pi - 2\beta$ radians in the circle of radius $4$ cm. Hence, using length of arc $= r\theta$ ,

\text{perimeter} = 3(6.283 - 1.854) + 4(6.283 - 1.287) = 33.3 \text{ cm}.

Figure 5.6. The figure-eight formed by the two discs, with its perimeter drawn heavily.

Problem 5.2.

The moon subtends an angle of $31'$ at the earth, and its distance from the earth is $382\,100$ km. Find the diameter of the moon in kilometres.
An arc $AB$ of length $5$ cm is marked on a circle of radius $3$ cm. Find the area of the sector bounded by this arc and the radii to $A$ and $B$ .
A chord $AB$ of length $5.2$ cm subtends an angle of $120^\circ$ at the centre of a circle. Calculate the length of the arc $AB$ , the area of the sector containing the angle of $120^\circ$ , and the area of the minor segment cut off by $AB$ .

The Ratios of a General Angle

Since we now regard an angle as the measure of the rotation of a line about a fixed point, the size of an angle is unlimited, because the line can keep on rotating indefinitely. The six trigonometric ratios, however, have so far been given a meaning only for acute angles, since each is defined by an angle in a right-angled triangle. To use them for angles of any size they must be defined in a more general way.

The system of reference in which a general angle is measured is very similar to that used for polar coordinates. The point about which the line $OP$ rotates is the pole or origin $O$ , and the position from which the angle is measured is the initial line, the $x$ -axis. An angle formed when the line rotates anticlockwise is positive, while clockwise rotation gives a negative angle. The pair of Cartesian axes divides the plane into four quadrants, numbered $1$ , $2$ , $3$ and $4$ as in Figure 5.7.

Figure 5.7. The four quadrants, and the positive and negative senses of rotation.

As the line $OP$ rotates, $P$ moves round the first quadrant, where both its coordinates are positive. As $OP$ moves into the second quadrant its $x$ -coordinate becomes negative. In the third quadrant both coordinates of $P$ are negative, and in the fourth quadrant the $x$ -coordinate is positive and the $y$ -coordinate negative. The length $r$ of $OP$ , the radius vector, is always taken to be positive.

Definition 5.6 (Trigonometric Ratios of Any Angle).

If the line $OP$ has rotated through the angle $\theta$ from the positive $x$ -axis, and $P$ is the point $(x, y)$ with $OP = r$ , then

\sin\theta = \frac{y}{r}, \qquad \cos\theta = \frac{x}{r}, \qquad \tan\theta = \frac{y}{x},

and $\operatorname{cosec}\theta = \dfrac{r}{y}$ , $\sec\theta = \dfrac{r}{x}$ , $\cot\theta = \dfrac{x}{y}$ , each ratio being defined whenever its denominator is not zero.

For an acute angle these agree with the ratios of the right-angled triangle formed by $O$ , $P$ and the foot of the perpendicular from $P$ to the $x$ -axis.

The numerical values of the ratios can be found as follows. From any position of $P$ , the perpendicular from $P$ meeting the $x$ -axis at $Q$ forms a right-angled triangle $OPQ$ . The angle $POQ$ formed in this way is always acute, whatever the value of $\theta$ , and is called the associated acute angle, $\alpha$ . For a particular value of $\theta$ , $\alpha$ is the difference between $\theta$ and $180^\circ$ or $360^\circ$ , or a further multiple of $180^\circ$ for larger angles. For example

\begin{aligned} \theta &= 160^\circ: & \alpha &= 180^\circ - 160^\circ = 20^\circ, &\qquad \theta &= \tfrac{7\pi}{6}: & \alpha &= \tfrac{7\pi}{6} - \pi = \tfrac{\pi}{6}, \\ \theta &= 275^\circ: & \alpha &= 360^\circ - 275^\circ = 85^\circ, &\qquad \theta &= 517^\circ: & \alpha &= 540^\circ - 517^\circ = 23^\circ . \end{aligned}

Figure 5.8. The associated acute angle

\alpha

when

P

is in each of the four quadrants.

In the triangle $OPQ$ the lengths $PQ$ and $OQ$ are $r\sin\alpha$ and $r\cos\alpha$ , so the ratios of $\theta$ have the numerical values of the ratios of $\alpha$ , and their signs depend on the signs of $x$ and $y$ , that is on the quadrant into which $P$ has rotated.

In the first quadrant all six ratios are positive, and since $\theta$ is acute their values are those of $\theta$ itself. If $OP$ has rotated through more than a complete revolution, $\alpha = \theta - 360^\circ$ .
In the second quadrant $\alpha = 180^\circ - \theta$ . The sine ratio $\dfrac{y}{r}$ is positive, the cosine ratio $\dfrac{x}{r}$ is negative and the tangent ratio $\dfrac{y}{x}$ is negative, so $\sin\theta = +\sin\alpha$ , $\cos\theta = -\cos\alpha$ and $\tan\theta = -\tan\alpha$ .
In the third quadrant $\alpha = \theta - 180^\circ$ . The tangent ratio is positive while the sine and cosine ratios are both negative, so $\sin\theta = -\sin\alpha$ , $\cos\theta = -\cos\alpha$ and $\tan\theta = +\tan\alpha$ .
In the fourth quadrant $\alpha = 360^\circ - \theta$ . The cosine ratio is positive but the sine and tangent ratios are negative, so $\sin\theta = -\sin\alpha$ , $\cos\theta = +\cos\alpha$ and $\tan\theta = -\tan\alpha$ .

These results are summarised in the quadrant diagrams of Figure 5.9. The second shows in each quadrant the ratios which are positive there: all of them in the first, the sine in the second, the tangent in the third and the cosine in the fourth. The quadrant rule, together with the value of the associated acute angle, gives the value of any trigonometric ratio of any angle.

Figure 5.9. Quadrant diagrams for the signs of the sine, cosine and tangent.

Negative Angles

If $OP$ rotates clockwise, so that $P$ moves through the quadrants in the reverse order, 4th, 3rd, 2nd, 1st, then $\theta$ is negative. Every position of $OP$ can be reached either by anticlockwise or by clockwise rotation, and so corresponds to two different values of $\theta$ , one positive and one negative. For example, $\theta = +120^\circ$ and $\theta = -240^\circ$ give the same position, with $\alpha = 60^\circ$ in both cases, so the angles $+120^\circ$ and $-240^\circ$ have the same trigonometric ratios. $P$ is in the second quadrant, where only the sine ratio is positive, hence

\begin{aligned} \sin 120^\circ &= \sin(-240^\circ) = +\sin 60^\circ, \\ \cos 120^\circ &= \cos(-240^\circ) = -\cos 60^\circ, \\ \tan 120^\circ &= \tan(-240^\circ) = -\tan 60^\circ . \end{aligned}

Figure 5.10. The same position of

OP

reached by rotations of

+120^\circ

and

-240^\circ

Turning through $-\theta$ instead of $\theta$ reflects $P$ in the $x$ -axis, which changes the sign of $y$ and leaves $x$ unchanged. So for every angle

\sin(-\theta) = -\sin\theta, \qquad \cos(-\theta) = \cos\theta, \qquad \tan(-\theta) = -\tan\theta .

Example 5.7.

Find the sine, cosine and tangent of $243^\circ$ .

Here $\theta = 243^\circ$ and $\alpha = \theta - 180^\circ = 63^\circ$ . $P$ is in the third quadrant, where only the tangent ratio is positive, so

\sin 243^\circ = -\sin 63^\circ = -0.8910, \qquad \cos 243^\circ = -\cos 63^\circ = -0.4540, \qquad \tan 243^\circ = +\tan 63^\circ = 1.9626 .

Example 5.8.

If $\cos\theta = 0.866$ and $\tan\theta$ is negative, find $\sin\theta$ .

$\theta$ is in a quadrant where the cosine ratio is positive and the tangent ratio is negative, that is in the fourth quadrant, where $\alpha = 360^\circ - \theta$ . Now $\cos\alpha = 0.866$ gives $\alpha = 30^\circ$ , so $\theta = 330^\circ$ . In the fourth quadrant the sine ratio is negative, therefore

\sin\theta = -\sin\alpha = -\sin 30^\circ = -0.5 .

Example 5.9.

Given that $\tan\theta = 1$ and $-2\pi \leqslant \theta \leqslant 2\pi$ , give four possible values of $\theta$ .

The tangent ratio is positive in the first and third quadrants. When $\tan\alpha = 1$ , $\alpha = \dfrac{\pi}{4}$ , or $45^\circ$ . The range of values of $\theta$ is specified in radians, so the solution should also be given in radians:

\theta = \frac{\pi}{4}, \qquad \frac{5\pi}{4}, \qquad -\frac{3\pi}{4} \qquad\text{or}\qquad -\frac{7\pi}{4} .

In this example we are solving a simple trigonometric equation.

Example 5.10.

An angle $\theta$ has an associated acute angle of $53^\circ$ . Find $\sin\theta$ .

Only $\alpha$ is given, so $\theta$ could be in any of the four quadrants. In quadrants $1$ and $2$ , $\sin\theta = \sin 53^\circ = 0.7986$ , and in quadrants $3$ and $4$ , $\sin\theta = -\sin 53^\circ = -0.7986$ . Therefore $\sin\theta = \pm 0.7986$ .

Problem 5.3.

Find the sine, cosine and tangent of $300^\circ$ and of $-160^\circ$ in terms of the ratios of their associated acute angles, and evaluate them.
Within the range $-360^\circ \leqslant \theta \leqslant 360^\circ$ , give all the values of $\theta$ for which $\cos\theta = -0.5$ , and all those for which $\tan\theta = 1.2$ .
Find the smallest angle, positive or negative, for which $\cos\theta = 0.8$ and $\sin\theta$ is positive, and the smallest for which $\sin\theta = -0.6$ and $\tan\theta$ is negative.

Solving Triangles

Triangles are involved in many practical measurements, in surveying for instance. A triangle has three sides and three angles, and its size and shape can be specified by suitable data, such as three sides, two angles and a side, or two sides and their included angle. The third angle is not an independent item: the angles of a triangle add up to $180^\circ$ , so if two of them are known the third follows directly. Two sides and a non-included angle can sometimes give two different triangles. From any data sufficient to define a triangle the remaining sides and angles can be calculated. This is called solving the triangle, and it uses one of the formulae that relate the sides and angles of a triangle. The two used most frequently are the sine rule and the cosine rule.

In a triangle $ABC$ we use $A$ , $B$ and $C$ to denote the angles at the vertices $A$ , $B$ and $C$ , and $a$ , $b$ and $c$ to denote the sides opposite these vertices.

The Sine Rule

In a triangle $ABC$ ,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} .

We show this by taking $O$ and $R$ as the centre and the radius of the circle through $A$ , $B$ and $C$ , drawing the diameter $AD$ and joining $DB$ , as in Figure 5.11. Then $\angle ABD = 90^\circ$ , the angle in a semicircle, so in the right-angled triangle $ABD$ , whose hypotenuse $AD$ is $2R$ ,

c = 2R\sin D .

In diagram (i), $C = D$ , since they are angles in the same segment. In diagram (ii), $C = 180^\circ - D$ , since they are opposite angles of a cyclic quadrilateral, and so $\sin C = \sin D$ by the rule for the second quadrant. In both diagrams, therefore, $\sin C = \sin D$ , and hence

c = 2R\sin C, \qquad\text{that is}\qquad \frac{c}{\sin C} = 2R .

Similarly $\dfrac{b}{\sin B} = 2R$ and $\dfrac{a}{\sin A} = 2R$ , so

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R .

Figure 5.11. The circle through

A

B

and

C

, with the diameter

AD

Any pair of these three equal ratios gives an equation containing two sides and two angles. So the sine rule can be used to solve a triangle in which we know either two sides and one angle, or two angles and one side, provided that one given side is opposite a given angle.

Example 5.11.

In a triangle $ABC$ , $A = 73^\circ$ , $B = 49^\circ$ and $a = 12.2$ cm. Find $b$ and $c$ .

As $a$ , $A$ and $B$ are known, $b$ can be calculated using

\frac{b}{\sin B} = \frac{a}{\sin A}, \qquad b = \frac{12.2\sin 49^\circ}{\sin 73^\circ} = 9.63 \text{ cm}.

Also $C = 180^\circ - 73^\circ - 49^\circ = 58^\circ$ , so $c$ can be calculated using

\frac{c}{\sin C} = \frac{a}{\sin A}, \qquad c = \frac{12.2\sin 58^\circ}{\sin 73^\circ} = 10.82 \text{ cm}.

In this example the given data define one and only one triangle.

The Ambiguous Case

Sometimes, when two sides and one angle are specified, two different triangles can be found from the data. Suppose that we have to solve the triangle in which $A = 24^\circ$ , $c = 2.6$ cm and $a = 1.1$ cm. Knowing $a$ , $A$ and $c$ we can find $C$ using

\frac{a}{\sin A} = \frac{c}{\sin C}, \qquad \sin C = \frac{2.6\sin 24^\circ}{1.1} = 0.9614 .

In a triangle, $\sin C = 0.9614$ gives $C = 74^\circ$ or $C = 106^\circ$ , since the sine is positive in both the first and the second quadrants. We must check whether both are possible values for $C$ .

If $C = 74^\circ$ , then $A + C = 98^\circ$ , which is less than $180^\circ$ . So $74^\circ$ is a possible value for $C$ , corresponding to $B = 82^\circ$ .
If $C = 106^\circ$ , then $A + C = 130^\circ$ , which is also less than $180^\circ$ . So $106^\circ$ is a possible value too, corresponding to $B = 50^\circ$ .

So there are two possible triangles with the given data. This is known as the ambiguous case, and it is easily understood by attempting to construct the triangle from its specification: an arc of radius $1.1$ about $B$ cuts the line from $A$ in two places. Each position of $C$ corresponds to a different pair of values for $B$ and $b$ , and in each case the solution of the triangle is completed using $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ , which gives $b = 2.68$ cm and $b = 2.07$ cm.

Figure 5.12. The two positions

C_1

and

C_2

of the vertex

C

in the ambiguous case.

There are not always two possible triangles when one angle and two sides are given, as the next example shows.

Example 5.12.

If $A = 37^\circ$ , $a = 4.59$ cm and $c = 2.1$ cm, show that there is only one possible triangle $ABC$ , and find its remaining angles.

Using $\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$ ,

\sin C = \frac{2.1\sin 37^\circ}{4.59} = 0.2753, \qquad\text{so}\qquad C = 16^\circ \ \text{or}\ C = 164^\circ .

If $C = 16^\circ$ and $A = 37^\circ$ , then $A + C = 53^\circ$ , which is less than $180^\circ$ , so $16^\circ$ is a possible value for $C$ , corresponding to $B = 127^\circ$ . If $C = 164^\circ$ , then $A + C = 201^\circ$ , which is more than $180^\circ$ , so $164^\circ$ is not a possible value. Hence there is only one triangle defined by the given data, and its other angles are $C = 16^\circ$ and $B = 127^\circ$ .

Problem 5.4.

In each part the data refer to a triangle in the standard notation.

$B = 35^\circ$ , $a = 2.7$ cm and $b = 5.1$ cm; find $A$ .
$b = 3.8$ cm, $A = 25^\circ$ and $a = 1.8$ cm; find the possible values of $C$ .
In a triangle $PQR$ the angle $PQR$ is $30^\circ$ and the angle $QPR$ is $\theta$ . Show that $\sin\theta = \dfrac{p}{2q}$ .

The Cosine Rule

The sine rule can be applied only when we know an angle opposite a given side. It cannot be used, for instance, when two sides and the included angle are given. Such cases are solved using the cosine rule,

a^2 = b^2 + c^2 - 2bc\cos A .

We show this by placing the triangle $ABC$ on Cartesian axes with $A$ at the origin and $AB$ along the positive $x$ -axis, as in Figure 5.13. In diagram (i) the coordinates of $B$ and $C$ are $(c, 0)$ and $(b\cos A, b\sin A)$ . In diagram (ii), where $A$ is obtuse, they are $(c, 0)$ and $\bigl(-b\cos(180^\circ - A), b\sin(180^\circ - A)\bigr)$ , which by the rule for the second quadrant is again $(b\cos A, b\sin A)$ . So by the length of the line joining two points,

a^2 = (b\sin A)^2 + (b\cos A - c)^2 = b^2\sin^2 A + b^2\cos^2 A - 2bc\cos A + c^2,

where $\sin^2 A$ means $(\sin A)^2$ . Since $C$ is at a distance $b$ from the origin, $(b\cos A)^2 + (b\sin A)^2 = b^2$ , and therefore

a^2 = b^2 + c^2 - 2bc\cos A .

Similarly it can be shown that

b^2 = c^2 + a^2 - 2ca\cos B, \qquad c^2 = a^2 + b^2 - 2ab\cos C .

Figure 5.13. The triangle

ABC

placed on axes, with

A

acute and with

A

obtuse.

The calculation involved in using the cosine rule is less straightforward than that required for the sine rule, so the cosine rule is used only when the sine rule is inapplicable.

Example 5.13.

In a triangle $ABC$ , $a = 17.5$ cm, $b = 8.4$ cm and $c = 11.9$ cm. Find the largest angle.

The longest side is opposite the largest angle, so we must find $A$ . As $a$ , $b$ and $c$ are given we use the cosine rule, rearranged so that $A$ can be found conveniently:

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{8.4^2 + 11.9^2 - 17.5^2}{2(8.4)(11.9)} = -0.4706 .

Hence $A = 118.1^\circ$ , which is obtuse because $\cos A$ is negative.

When the solution of a triangle requires the cosine rule first, the remaining sides and angles can then be found using the sine rule.

Example 5.14.

Solve the triangle $PQR$ given that $p = 12.1$ cm, $q = 7.3$ cm and $R = 37.5^\circ$ .

We first use

r^2 = p^2 + q^2 - 2pq\cos R, \qquad\text{which gives}\qquad r = 7.72 \text{ cm}.

Now the sine rule can be used, since we know $r$ , $R$ and $q$ :

\frac{q}{\sin Q} = \frac{r}{\sin R}, \qquad \sin Q = \frac{7.3\sin 37.5^\circ}{7.72} = 0.5756 .

Hence $Q = 35.1^\circ$ . $Q$ cannot be obtuse, as $q$ is not the longest side of the triangle. Finally $P = 180^\circ - Q - R = 107.4^\circ$ .

Problem 5.5.

Solve the triangle in which $a = 4$ , $b = 7$ and $c = 5$ .
Using $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ , show that $A$ is acute if $a^2 < b^2 + c^2$ and obtuse if $a^2 > b^2 + c^2$ , and check that taking $A = 90^\circ$ gives the result of Pythagoras.
Find the angles of a triangle whose sides are in the ratio $2 : 3 : 4$ .

The Graphs of the Circular Functions

There is a single value for each trigonometric ratio of any angle, so the mappings $\theta \mapsto \sin\theta$ , $\theta \mapsto \cos\theta$ , and so on, are functions, and we can plot graphs showing how each trigonometric function behaves as $\theta$ varies. The graphs of the sine, cosine and tangent are particularly important.

The Sine Function

Figure 5.14. The graph of

f(\theta) = \sin\theta

The graph of $f : \theta \mapsto \sin\theta$ , $\theta \in \RR$ , shows that the sine function has the following characteristics.

It is continuous: its graph has no breaks.
Its range is $-1 \leqslant \sin\theta \leqslant 1$ .
The shape of the graph from $\theta = 0$ to $\theta = 2\pi$ is repeated for each further complete revolution.

Definition 5.15 (Periodic Function).

A function whose graph repeats a pattern is periodic, or cyclic. The width of the repeating pattern, measured on the horizontal axis, is the period of the function: it is the smallest positive number $p$ for which $f(\theta + p) = f(\theta)$ for every $\theta$ .

So $\theta \mapsto \sin\theta$ is a periodic function with a period of $2\pi$ , a maximum value of $1$ and a minimum value of $-1$ . A graph of this shape is known as a sine wave. The amplitude is half the distance between the maximum and minimum values; for $\sin\theta$ its value is $1$ .

The Cosine Function

Figure 5.15. The graph of

f(\theta) = \cos\theta

, with the sine curve dashed.

The characteristics of the graph of $f : \theta \mapsto \cos\theta$ , $\theta \in \RR$ , are as follows.

It is continuous.
It lies entirely within the range $-1 \leqslant \cos\theta \leqslant 1$ .
It is periodic with a period of $2\pi$ .
It has the same shape as the sine graph, but is displaced a distance $\dfrac{\pi}{2}$ to the left on the horizontal axis. Such a displacement is known as a phase difference, or phase shift.

So $\theta \mapsto \cos\theta$ is a cyclic function with period $2\pi$ and values from $-1$ to $1$ .

The Tangent Function

Figure 5.16. The graph of

f(\theta) = \tan\theta

The behaviour of the tangent function $f : \theta \mapsto \tan\theta$ is different from that of the sine and cosine functions in several respects.

It is not continuous, being undefined when $\theta = \pm\dfrac{\pi}{2}, \pm\dfrac{3\pi}{2}, \pm\dfrac{5\pi}{2}, \ldots$ , where the lines drawn dashed in Figure 5.16 are asymptotes to the curve.
The range of possible values of $\tan\theta$ is unlimited.
The tangent function is periodic, but its period is $\pi$ , not $2\pi$ as for the sine and cosine.

Special Values

It is useful to note the angles whose trigonometric ratios have the values $0$ and $\pm 1$ , and the angles at which $\tan\theta$ is undefined. Reference to the graphs shows that, for $n \in \ZZ$ ,

\begin{aligned} \sin\theta &= 0 && \text{when } \theta = \ldots, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi, \ldots, &&\text{that is } \theta = n\pi, \\ \sin\theta &= 1 && \text{when } \theta = \ldots, -\tfrac{3\pi}{2}, \tfrac{\pi}{2}, \tfrac{5\pi}{2}, \ldots, &&\text{that is } \theta = 2n\pi + \tfrac{\pi}{2}, \\ \sin\theta &= -1 && \text{when } \theta = \ldots, -\tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{7\pi}{2}, \ldots, &&\text{that is } \theta = 2n\pi - \tfrac{\pi}{2}, \\ \cos\theta &= 0 && \text{when } \theta = \ldots, -\tfrac{\pi}{2}, \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \ldots, &&\text{that is } \theta = (2n + 1)\tfrac{\pi}{2}, \\ \cos\theta &= 1 && \text{when } \theta = \ldots, -2\pi, 0, 2\pi, 4\pi, \ldots, &&\text{that is } \theta = 2n\pi, \\ \cos\theta &= -1 && \text{when } \theta = \ldots, -\pi, \pi, 3\pi, 5\pi, \ldots, &&\text{that is } \theta = (2n + 1)\pi, \\ \tan\theta &= 0 && \text{when } \theta = \ldots, -\pi, 0, \pi, 2\pi, \ldots, &&\text{that is } \theta = n\pi, \end{aligned}

and $\tan\theta$ is undefined, its graph going off to $\pm\infty$ , when $\theta = (2n + 1)\dfrac{\pi}{2}$ .

Remark.

Throughout, $2n$ stands for any even integer and $2n + 1$ for any odd integer, provided that $n \in \ZZ$ .

The Reciprocal Ratios

The three ratios $\sin\theta$ , $\cos\theta$ and $\tan\theta$ are used much more frequently than their reciprocals $\operatorname{cosec}\theta$ , $\sec\theta$ and $\cot\theta$ , but the reciprocal ratios must not be overlooked. The graph of $f(\theta) = \operatorname{cosec}\theta$ can be drawn without any table of values, simply by observing the graph of $f(\theta) = \sin\theta$ and using the following properties of any expression and its reciprocal.

As an expression approaches zero its reciprocal becomes numerically large without limit, and as an expression becomes numerically large without limit its reciprocal approaches zero.
The reciprocal of $1$ is $1$ , and the reciprocal of $-1$ is $-1$ .
Where an expression has a maximum value its reciprocal has a minimum value, and conversely.
Where an expression is increasing its reciprocal is decreasing, and conversely.
An expression and its reciprocal have the same sign.

These properties are reasonable enough to accept without detailed analysis at this stage.

Figure 5.17. The graph of

f(\theta) = \operatorname{cosec}\theta

for

-\pi \leqslant \theta \leqslant 2\pi

, with the sine curve dashed.

In the same way the graphs of $f(\theta) = \sec\theta$ and $f(\theta) = \cot\theta$ can be deduced from those of $\cos\theta$ and $\tan\theta$ .

Figure 5.18. The graphs of

\sec\theta

and

\cot\theta

, each drawn over the curve it is the reciprocal of.

Inverse Circular Functions

The function $f : x \mapsto \sin x$ is a many-one mapping for the domain $x \in \RR$ , and so it does not have an inverse function. However, if the domain is redefined as $-\dfrac{\pi}{2} \leqslant x \leqslant \dfrac{\pi}{2}$ , the function $f : x \mapsto \sin x$ is a one-one mapping, and now it does have an inverse. This inverse sine function is denoted by $\arcsin$ or $\sin^{-1}$ . Thus

\text{if } f : x \mapsto \sin x, \ -\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}, \qquad\text{then}\qquad f^{-1} : x \mapsto \arcsin x, \ -1 \leqslant x \leqslant 1 .

For $f : x \mapsto \sin x$ the input is an angle and the output is a number. So for the inverse function $f^{-1} : x \mapsto \arcsin x$ the input is a number and the output is an angle, and $\arcsin x$ means “the angle whose sine is $x$ ”.

Similarly, if $f : x \mapsto \cos x$ , $0 \leqslant x \leqslant \pi$ , then $f^{-1}$ exists and is denoted by $\arccos$ or $\cos^{-1}$ , where $\arccos x$ means “the angle whose cosine is $x$ ”:

\text{if } f : x \mapsto \cos x, \ 0 \leqslant x \leqslant \pi, \qquad\text{then}\qquad f^{-1} : x \mapsto \arccos x, \ -1 \leqslant x \leqslant 1 .

Further, if $f : x \mapsto \tan x$ for $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ , the inverse function $f^{-1}$ exists and is written $\arctan$ or $\tan^{-1}$ , where $\arctan x$ means “the angle whose tangent is $x$ ”. The domain of $x \mapsto \arctan x$ is $x \in \RR$ . In the same way $\operatorname{arccot} x$ is the angle whose cotangent is $x$ , and for positive $x$ it is $\arctan\dfrac{1}{x}$ .

Figure 5.19. The restricted sine, cosine and tangent functions and their inverses, reflected in the line

y = x

Remark (A warning about notation).

If the notation $\sin^{-1}$ is adopted, it is most important to appreciate that $\sin^{-1} x$ is not the same as $\dfrac{1}{\sin x}$ .

Common Trigonometric Ratios

The angles $30^\circ$ , $45^\circ$ and $60^\circ$ , and the angles for which these are the associated acute angles, such as $150^\circ$ , $225^\circ$ and $300^\circ$ , are used frequently, so their trigonometric ratios are well worth noting.

Consider first an equilateral triangle $ABC$ which is bisected by the line $AD$ . In the triangle $BAD$ the angle $B$ is $60^\circ$ , or $\dfrac{\pi}{3}$ , since the triangle $ABC$ is equilateral, and the angle at $A$ is $30^\circ$ , or $\dfrac{\pi}{6}$ , since the angle $BAC$ is bisected. If $AB = 2$ units then $BD = 1$ unit, and $AD = \sqrt3$ units by Pythagoras. Therefore

\begin{aligned} \sin\frac{\pi}{6} &= \sin 30^\circ = \frac12, & \sin\frac{\pi}{3} &= \sin 60^\circ = \frac{\sqrt3}{2}, \\ \cos\frac{\pi}{6} &= \cos 30^\circ = \frac{\sqrt3}{2}, & \cos\frac{\pi}{3} &= \cos 60^\circ = \frac12, \\ \tan\frac{\pi}{6} &= \tan 30^\circ = \frac{1}{\sqrt3}, & \tan\frac{\pi}{3} &= \tan 60^\circ = \sqrt3 . \end{aligned}

Alternatively we can say

\begin{aligned} \arcsin\frac12 &= \frac{\pi}{6}, & \arccos\frac{\sqrt3}{2} &= \frac{\pi}{6}, & \arctan\frac{1}{\sqrt3} &= \frac{\pi}{6}, \\ \arcsin\frac{\sqrt3}{2} &= \frac{\pi}{3}, & \arccos\frac12 &= \frac{\pi}{3}, & \arctan\sqrt3 &= \frac{\pi}{3} . \end{aligned}

Now consider a triangle $ABC$ in which $AB = BC$ and the angle $B$ is a right angle, so that the angles at $A$ and $C$ are each $45^\circ$ . If $AB = BC = 1$ unit, then $AC = \sqrt2$ units by Pythagoras. Therefore

\sin\frac{\pi}{4} = \sin 45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}, \qquad \cos\frac{\pi}{4} = \cos 45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}, \qquad \tan\frac{\pi}{4} = \tan 45^\circ = 1,

and $\arcsin\dfrac{1}{\sqrt2} = \dfrac{\pi}{4} = \arccos\dfrac{1}{\sqrt2}$ , $\arctan 1 = \dfrac{\pi}{4}$ .

Figure 5.20. The triangles which give the ratios of

30^\circ

60^\circ

and

45^\circ

Complementary Angles

Definition 5.16 (Complementary Angles).

If the sum of two acute angles is $90^\circ$ , or $\dfrac{\pi}{2}$ , they are complementary, and each is the complement of the other.

Consider a right-angled triangle $ABC$ containing the angles $\alpha$ and $\beta$ , as in Figure 5.21. Then

\sin\alpha = \frac{a}{c} = \cos\beta, \qquad \cos\alpha = \frac{b}{c} = \sin\beta, \qquad \tan\alpha = \frac{a}{b} = \cot\beta, \qquad \cot\alpha = \frac{b}{a} = \tan\beta .

Figure 5.21. A right-angled triangle containing two complementary angles

\alpha

and

\beta

But $\alpha$ and $\beta$ are complementary, so we have shown that the sine of an angle is the cosine of its complement, and the tangent of an angle is the cotangent of its complement. Because of this property the sine and cosine of an angle are called complementary ratios, and similarly the tangent and cotangent are complementary ratios.

The same relationships hold for angles of any size. Reflecting $OP$ in the line $y = x$ exchanges the coordinates $x$ and $y$ of $P$ , and turns the angle $\theta$ into $\dfrac{\pi}{2} - \theta$ , so for every angle

\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right), \qquad \cos\theta = \sin\left(\frac{\pi}{2} - \theta\right), \qquad \tan\theta = \cot\left(\frac{\pi}{2} - \theta\right) .

Trigonometric Equations

Definition 5.17 (Trigonometric Equation).

An equation in which at least one term contains a trigonometric ratio is a trigonometric equation. Solving it means finding the angle or angles for which it is true.

Consider the simple equation $\sin\theta = 0$ . Referring to the graph of the sine function, we see that $\sin\theta = 0$ when $\theta$ is any multiple of $\pi$ , that is when $\theta = n\pi$ , where $n \in \ZZ$ . The full, or general, solution of this equation is the infinite set of angles $\theta = n\pi$ , or $\theta = 180n^\circ$ .

Sometimes it is necessary to extract certain values of $\theta$ from the infinite set. Solving the equation $\sin\theta = 0$ for $-\pi \leqslant \theta \leqslant \pi$ , for instance, gives the finite solution set $\theta = -\pi, 0, \pi$ .

There are two basic approaches to finding the solution of a trigonometric equation. One of them was used above, and refers to the graph of the appropriate circular function; it is usually best for sines and cosines with the values $\pm 1$ and $0$ , and for tangents which are zero or undefined. Alternatively, the position of the rotating line $OP$ in the appropriate quadrants can lead to a clear solution. In all cases the first step is to find the principal solution, which is the principal value, PV, of $\theta$ .

Principal Values

Figure 5.22. The intervals in which each value of the sine, cosine and tangent occurs exactly once.

From the graph of the sine function, every possible value of $\sin\theta$ occurs once and only once in the interval $-\dfrac{\pi}{2} \leqslant \theta \leqslant \dfrac{\pi}{2}$ . So any equation $\sin\theta = s$ , with $-1 \leqslant s \leqslant 1$ , has one and only one solution in this interval, and this is the principal value of $\theta$ , in either the first or the fourth quadrant. For example, if $\sin\theta = \frac12$ the principal solution is $\theta = \dfrac{\pi}{6}$ , and if $\sin\theta = -\frac12$ it is $\theta = -\dfrac{\pi}{6}$ .
Every possible value of $\cos\theta$ occurs once and only once in the interval $0 \leqslant \theta \leqslant \pi$ , so there is one and only one solution of $\cos\theta = c$ in this interval. This is the principal value of $\theta$ , in either the first or the second quadrant. If $\cos\theta = \frac12$ the principal solution is $\theta = \dfrac{\pi}{3}$ , and if $\cos\theta = -\frac12$ it is $\theta = \dfrac{2\pi}{3}$ .
Every possible value of $\tan\theta$ occurs once and only once for angles in the interval $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$ , so one and only one solution of $\tan\theta = t$ is in this interval, and it is the principal value of $\theta$ , in the first or the fourth quadrant. If $\tan\theta = 1$ the principal solution is $\theta = \dfrac{\pi}{4}$ , and if $\tan\theta = -1$ it is $\theta = -\dfrac{\pi}{4}$ .

These intervals are the ranges of the inverse functions, so the principal values of the solutions of $\sin\theta = s$ , $\cos\theta = c$ and $\tan\theta = t$ are $\arcsin s$ , $\arccos c$ and $\arctan t$ .

Secondary Values

Having found the principal value of the solution of a trigonometric equation, we usually find a second angle with the same trigonometric ratio in the interval $-\pi < \theta \leqslant \pi$ . This solution lies in a different quadrant and is called the secondary value, SV, of $\theta$ , or the secondary solution of the equation.

If $\sin\theta = \frac12$ , the secondary solution is in the second quadrant, where the sine ratio is also positive, and it is $\theta = \dfrac{5\pi}{6}$ . If $\sin\theta = -\frac12$ , the secondary solution is in the third quadrant, where the sine ratio is also negative, and it is $\theta = -\dfrac{5\pi}{6}$ .
If $\cos\theta = \frac12$ , the secondary value is in the fourth quadrant and is $\theta = -\dfrac{\pi}{3}$ . If $\cos\theta = -\frac12$ , the secondary value is in the third quadrant and is $\theta = -\dfrac{2\pi}{3}$ . For an equation of the form $\cos\theta = c$ , $\text{SV} = -\text{PV}$ .
If $\tan\theta = 1$ , the secondary solution is in the third quadrant and is $\theta = -\dfrac{3\pi}{4}$ . If $\tan\theta = -1$ , the secondary value is in the second quadrant and is $\theta = \dfrac{3\pi}{4}$ .

Figure 5.23. Principal and secondary values for six equations.

Problem 5.6.

Determine the principal solutions of $\sin\theta = -\dfrac{\sqrt3}{2}$ , $\cos\theta = -\dfrac{1}{\sqrt2}$ and $\tan\theta = -\dfrac{\sqrt3}{3}$ .
Find the principal and secondary solutions of $\sin\theta = \dfrac{1}{\sqrt2}$ , $\cos\theta = -\dfrac{\sqrt3}{2}$ and $\tan\theta = -\sqrt3$ .
By referring to the graphs of the appropriate circular functions, explain why the equations $\sin\theta = -1$ and $\cos\theta = 1$ have no secondary solution.

Solutions in a Specified Range

In solving a trigonometric equation in a specified range we first find the principal angle and the secondary angle, except in those cases where there is no secondary angle. A quadrant diagram can then be drawn showing the two solution positions, and any angle measured from the positive $x$ -axis to either of the solution positions is a solution of the equation.

Example 5.18.

Solve the equation $\sin\theta = 0.4$ within the interval $-360^\circ \leqslant \theta \leqslant 360^\circ$ .

The principal solution is $\theta = 23.58^\circ$ . The secondary solution is in the second quadrant, since the sine ratio is positive in the first and second quadrants, and it is $\theta = 156.42^\circ$ . Therefore the solutions within the specified interval are

\theta = -336.42^\circ, \quad -203.58^\circ, \quad 23.58^\circ, \quad 156.42^\circ .

Figure 5.24. The four solutions of

\sin\theta = 0.4

between

-360^\circ

and

360^\circ

Example 5.19.

Solve the equation $\tan\theta = -\dfrac{1}{\sqrt3}$ in the interval $0 \leqslant \theta \leqslant 2\pi$ .

If $\tan\theta = -\dfrac{1}{\sqrt3}$ , the principal solution is in the fourth quadrant and the secondary solution is in the second quadrant: the PV is $-\dfrac{\pi}{6}$ and the SV is $\dfrac{5\pi}{6}$ . Within the specified interval the solution set is

\theta = \frac{5\pi}{6}, \quad \frac{11\pi}{6} .

As here, the principal value is not always included in the solution set.

Example 5.20.

Find the angles in the interval $-360^\circ \leqslant \theta \leqslant 0$ which satisfy the equation $\cos\theta = 0.7$ .

Since $\cos\theta$ is positive, the principal solution is in the first quadrant and the secondary solution is in the fourth: the PV is $45.57^\circ$ and the SV is $-45.57^\circ$ . In the interval $-360^\circ \leqslant \theta \leqslant 0$ the solution set is $\theta = -314.43^\circ, -45.57^\circ$ .

Example 5.21.

Solve, within the interval $0 \leqslant \theta \leqslant 360^\circ$ , the equation $\sin\theta + 3\sin\theta\cos\theta = 0$ .

First the equation must be factorised:

\sin\theta\,(1 + 3\cos\theta) = 0 .

Therefore either $\sin\theta = 0$ or $\cos\theta = -\frac13$ .

Referring to the sine graph, $\sin\theta = 0$ gives $\theta = 0, 180^\circ, 360^\circ$ .
For $\cos\theta = -\frac13$ the principal solution is in the second quadrant and the secondary solution in the third: the PV is $109.47^\circ$ and the SV is $-109.47^\circ$ . Within the specified range these give $\theta = 109.47^\circ, 250.53^\circ$ .

So the complete solution set from $0$ to $360^\circ$ is $\theta = 0, 109.47^\circ, 180^\circ, 250.53^\circ, 360^\circ$ .

Although the solutions of $\sin\theta = 0$ could conveniently be expressed in radians, degrees are used because the range is specified in degrees. Units must not be mixed in any one example.

Problem 5.7.

Solve the following equations for angles in the interval $0 \leqslant \theta \leqslant 2\pi$ .

$\sqrt3\tan\theta = 2\sin\theta$ ;
$2\sin\theta\cos\theta + \sin\theta = 0$ ;
$4\cos\theta = \cos\theta\operatorname{cosec}\theta$ .

General Solutions

Definition 5.22 (General Solution).

The general solution of a trigonometric equation is an expression which represents all the angles which satisfy the equation, that is an infinite set of angles.

In looking for a general solution we use the graphs of the circular functions, the period of each circular function, and the principal solution together with, except when the tangent ratio is involved, the secondary solution.

Consider the equation $\sin\theta = s$ , where $-1 \leqslant s \leqslant 1$ . The period $2\pi$ of the sine function is covered by the interval $-\pi < \theta \leqslant \pi$ , which includes both the PV and the SV of $\theta$ . So by adding or subtracting any multiple of $2\pi$ to either the PV or the SV we get another angle with the same sine. Thus the complete solution of $\sin\theta = s$ is

\theta = \text{PV} + 2n\pi \quad\text{or}\quad \theta = \text{SV} + 2n\pi, \qquad n \in \ZZ,

or, in degrees, $\theta = \text{PV} + 360n^\circ$ or $\theta = \text{SV} + 360n^\circ$ .

A similar situation arises for the equation $\cos\theta = c$ , because both the PV and the SV of the cosine lie within one period, which is again $2\pi$ . So the complete solution of $\cos\theta = c$ is also given by adding multiples of $2\pi$ to either the PV or the SV. Remembering that for cosines the PV and the SV are equal in value but opposite in sign, the general solution of $\cos\theta = c$ can be given in the form

\theta = \pm\text{PV} + 2n\pi \quad\text{or}\quad \theta = \pm\text{PV} + 360n^\circ, \qquad n \in \ZZ .

For the equation $\tan\theta = t$ , only the principal value is included in the complete period $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$ . All further angles with the same tangent are given by adding multiples of $\pi$ , the period, to the PV:

\theta = \text{PV} + n\pi, \qquad n \in \ZZ .

Example 5.23.

Find the general solution of each of the equations $\tan\theta = 1$ and $\tan\theta = -\sqrt3$ .

The principal solution of $\tan\theta = 1$ is $\theta = \dfrac{\pi}{4}$ , so the general solution is $\theta = \dfrac{\pi}{4} + n\pi$ , where $n \in \ZZ$ .

If $\tan\theta = -\sqrt3$ , the principal solution is $\theta = -\dfrac{\pi}{3}$ , and the general solution is therefore $\theta = n\pi - \dfrac{\pi}{3}$ .

Example 5.24.

Find the general solution of each of the equations $\cos\theta = \dfrac{1}{\sqrt2}$ and $\cos\theta = -\dfrac12$ .

The principal solution of $\cos\theta = \dfrac{1}{\sqrt2}$ is $\theta = \dfrac{\pi}{4}$ , so the general solution is $\theta = 2n\pi \pm \dfrac{\pi}{4}$ .

When $\cos\theta = -\frac12$ , the principal value of $\theta$ is $\dfrac{2\pi}{3}$ , and the general solution is $\theta = 2n\pi \pm \dfrac{2\pi}{3}$ .

Example 5.25.

Find the general solution set of the equation $\sin\theta = \frac12$ .

The principal value of $\theta$ for which $\sin\theta = \frac12$ is $\dfrac{\pi}{6}$ , and the secondary value, in the second quadrant, is $\dfrac{5\pi}{6}$ . So the general solution set includes

\theta = \frac{\pi}{6} + 2n\pi \qquad\text{and}\qquad \theta = \frac{5\pi}{6} + 2n\pi .

Remark.

There is a way of combining the two parts of the general solution of $\sin\theta = s$ into one formula,

\theta = (-1)^n\,\text{PV} + n\pi, \qquad n \in \ZZ .

When $n$ is even this is $\text{PV} + 2k\pi$ , and when $n$ is odd it is $\pi - \text{PV} + 2k\pi$ , which is the secondary value plus a multiple of $2\pi$ . So when $\sin\theta = \frac12$ , $\theta = (-1)^n\dfrac{\pi}{6} + n\pi$ . Either form of the general solution may be used.

Example 5.26.

Find the general solution of the equation $4\sin\theta\,(2\tan\theta + 3) + 6\tan\theta + 9 = 0$ .

First we simplify and factorise the equation:

4\sin\theta\,(2\tan\theta + 3) + 3(2\tan\theta + 3) = 0, \qquad (4\sin\theta + 3)(2\tan\theta + 3) = 0 .

So either $\sin\theta = -\frac34$ or $\tan\theta = -\frac32$ .

For $\sin\theta = -\frac34$ the principal solution is $\theta = -48.59^\circ$ , and the secondary solution, in the third quadrant, is $\theta = -131.41^\circ$ . Hence the general solution is $\theta = -48.59^\circ + 360n^\circ$ or $\theta = -131.41^\circ + 360n^\circ$ .
For $\tan\theta = -\frac32$ the principal solution is the only one we need, and it is $\theta = -56.31^\circ$ . The general solution is then $\theta = -56.31^\circ + 180n^\circ$ .

Combining these results, the general solution set of the given equation is

\theta = -48.59^\circ + 360n^\circ, \qquad \theta = -131.41^\circ + 360n^\circ, \qquad \theta = -56.31^\circ + 180n^\circ .

Problem 5.8.

Find the general solution of each of the following equations.

$\sin\theta = -\dfrac{\sqrt3}{2}$ ;
$\cos\theta = 0$ ;
$\cos\theta = 0.371$ .

Multiple Angles

Equations are frequently met in which the angle involved is a multiple of $\theta$ , such as $\cos 2\theta = \frac12$ or $\tan 3\theta = -2$ . Such equations are solved by determining first the necessary values of the multiple angle and then, by division, the corresponding values of $\theta$ .

Example 5.27.

Find the general solution of the equation $\cos 2\theta = \frac12$ .

Let $2\theta = \varphi$ , so that $\cos\varphi = \frac12$ . The principal value of $\varphi$ is $\dfrac{\pi}{3}$ , so the general solution for $\varphi$ is $\varphi = \pm\dfrac{\pi}{3} + 2n\pi$ , that is

2\theta = \pm\frac{\pi}{3} + 2n\pi, \qquad\text{hence}\qquad \theta = \pm\frac{\pi}{6} + n\pi .

Example 5.28.

Find the angles within the range $-180^\circ \leqslant \theta \leqslant 180^\circ$ which satisfy the equation $\tan 3\theta = -2$ .

Let $3\theta = \varphi$ , so that $\tan\varphi = -2$ . The principal value is in the fourth quadrant, $\varphi = -63.43^\circ$ , and the secondary value is in the second quadrant, $\varphi = 116.57^\circ$ . Values of $\theta$ are required in the range $-180^\circ \leqslant \theta \leqslant 180^\circ$ , and $\varphi = 3\theta$ , so we need the values of $\varphi$ in the range $-540^\circ \leqslant \varphi \leqslant 540^\circ$ . These are

\varphi = -423.43^\circ, \quad -243.43^\circ, \quad -63.43^\circ, \quad 116.57^\circ, \quad 296.57^\circ, \quad 476.57^\circ,

and therefore

\theta = -141.14^\circ, \quad -81.14^\circ, \quad -21.14^\circ, \quad 38.86^\circ, \quad 98.86^\circ, \quad 158.86^\circ .

Alternatively, quoting the general solution for $\varphi$ gives $\varphi = -63.43^\circ + 180n^\circ$ , so that $\theta = -21.14^\circ + 60n^\circ$ . Giving $n$ the values $-2, -1, 0, 1, 2, 3$ , which cover the required range, gives the same values of $\theta$ .

Example 5.29.

Find the solutions of the equation $\sin\dfrac{\theta}{2} = 0.6$ for values of $\theta$ between $0$ and $360^\circ$ .

Let $\dfrac{\theta}{2} = \varphi$ , so that $\sin\varphi = 0.6$ . The principal value of $\varphi$ is $36.87^\circ$ , and the secondary value, in the second quadrant, is $143.13^\circ$ . The required range of values of $\theta$ is from $0$ to $360^\circ$ , so the range of values of $\varphi$ is from $0$ to $180^\circ$ , which gives $\varphi = 36.87^\circ, 143.13^\circ$ . Therefore $\theta = 73.74^\circ, 286.26^\circ$ .

Problem 5.9.

Solve the equations $\tan 2\theta = 1$ and $\sin 3\theta = 0.7$ within the interval $0 \leqslant \theta \leqslant 360^\circ$ .
Find the general solution of the equation $\cos 2\theta = 0.63$ .

The Equation $\cos A = \cos B$

This type of equation can be solved very neatly as follows. Let $\cos A = \cos B = c$ , where $-1 \leqslant c \leqslant 1$ . In general there are two solution positions for $\cos B = c$ , $OP_1$ and $OP_2$ , and the set of angles represented by $OP_1$ and $OP_2$ is $2n\pi \pm B$ . But we also know that $\cos A = c$ , so $OP_1$ and $OP_2$ together represent all possible values of $A$ . Thus

\cos A = \cos B \qquad\text{gives}\qquad A = 2n\pi \pm B,

that is, the values of $A$ are the general solution set for $B$ . The same argument for tangents and sines gives

\tan A = \tan B \ \text{ gives } \ A = n\pi + B, \qquad \sin A = \sin B \ \text{ gives } \ A = 2n\pi + B \ \text{ or } \ A = (2n + 1)\pi - B .

Example 5.30.

Solve the equation $\cos 4\theta = \cos\theta$ .

Using the conclusion above,

4\theta = 2n\pi \pm \theta, \qquad\text{so}\qquad 5\theta = 2n\pi \quad\text{or}\quad 3\theta = 2n\pi,

and hence $\theta = \dfrac{2n\pi}{5}$ or $\theta = \dfrac{2n\pi}{3}$ .

Example 5.31.

Find the values in the range $0 \leqslant \theta \leqslant 360^\circ$ which satisfy the equation $\tan(3\theta - 40^\circ) = \tan\theta$ .

The general solution is

3\theta - 40^\circ = 180n^\circ + \theta, \qquad 2\theta = 180n^\circ + 40^\circ, \qquad \theta = 90n^\circ + 20^\circ .

For $0 \leqslant \theta \leqslant 360^\circ$ we let $n = 0, 1, 2, 3$ , giving $\theta = 20^\circ, 110^\circ, 200^\circ, 290^\circ$ .

This method can be used only for equations containing two terms involving the same trigonometric ratio. That situation can sometimes be arranged in an apparently unsuitable case, as in the next example.

Example 5.32.

Find the general solution of $\cos 3\theta = \sin\theta$ .

We know that $\sin\theta = \cos\left(\dfrac{\pi}{2} - \theta\right)$ , from complementary angles, so the equation can be written

\cos 3\theta = \cos\left(\frac{\pi}{2} - \theta\right), \qquad\text{so}\qquad 3\theta = 2n\pi \pm \left(\frac{\pi}{2} - \theta\right) .

Therefore either $4\theta = 2n\pi + \dfrac{\pi}{2}$ or $2\theta = 2n\pi - \dfrac{\pi}{2}$ , that is

\theta = \frac{n\pi}{2} + \frac{\pi}{8} \qquad\text{or}\qquad \theta = n\pi - \frac{\pi}{4} .

Problem 5.10.

Find the general solutions of the equations $\cos 4\theta = \cos 3\theta$ and $\sin 4\theta = \sin 3\theta$ .
Solve the equation $\cos(2\theta + 60^\circ) = \cos\theta$ , giving the values of $\theta$ from $-180^\circ$ to $180^\circ$ .

Graphs of Multiple and Compound Angles

Consider $f : \theta \mapsto \sin 2\theta$ . The following table gives pairs of corresponding values of $\theta$ and $f(\theta)$ .

$\theta$	$0$	$\frac{\pi}{4}$	$\frac{\pi}{2}$	$\frac{3\pi}{4}$	$\pi$	$\frac{5\pi}{4}$	$\frac{3\pi}{2}$	$\frac{7\pi}{4}$	$2\pi$
$2\theta$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$	$\frac{5\pi}{2}$	$3\pi$	$\frac{7\pi}{2}$	$4\pi$
$f(\theta)$	$0$	$1$	$0$	$-1$	$0$	$1$	$0$	$-1$	$0$

Figure 5.25. The graph of

f(\theta) = \sin 2\theta

, with the graph of

\sin\theta

dashed.

The following characteristics can be observed.

The function $\sin 2\theta$ is cyclic, and its period is $\pi$ , that is $\frac12 \times 2\pi$ .
Its range is $-1 \leqslant \sin 2\theta \leqslant 1$ .
Its shape is a sine wave.
Within the domain $0 \leqslant \theta \leqslant 2\pi$ there are two complete cycles of the curve, compared with only one for the basic function $\theta \mapsto \sin\theta$ : the complete cycle appears with twice the frequency.

When the same investigation is carried out on $\theta \mapsto \sin 3\theta$ , we find that the function is cyclic with period $\dfrac{2\pi}{3}$ , so that three complete cycles occur between $0$ and $2\pi$ . In general the graph of $\theta \mapsto \sin k\theta$ , for $k > 0$ , is a sine wave with period $\dfrac{2\pi}{k}$ and a frequency $k$ times that of $\theta \mapsto \sin\theta$ . We show this by noting that

\sin k\left(\theta + \frac{2\pi}{k}\right) = \sin(k\theta + 2\pi) = \sin k\theta,

so that the graph repeats after a width of $\dfrac{2\pi}{k}$ , while as $\theta$ runs through any interval of that width, $k\theta$ runs through an interval of width $2\pi$ and so through one complete sine wave. Similar properties hold for $\theta \mapsto \cos k\theta$ , whose period is $\dfrac{2\pi}{k}$ , and for $\theta \mapsto \tan k\theta$ , whose period is $\dfrac{\pi}{k}$ . For example, the period of $\tan 2\theta$ is $\dfrac{\pi}{2}$ and its frequency is $2$ ; the period of $\cos 4\theta$ is $\dfrac{\pi}{2}$ and its frequency is $4$ ; the period of $\sin\dfrac{\theta}{2}$ is $4\pi$ and its frequency is $\frac12$ .

Figure 5.26. The graphs of

\tan 2\theta

\cos 4\theta

and

\sin\frac12\theta

Now consider the function $f(\theta) = \cos(\theta - \alpha)$ . This is clearly a cosine function, but

$f(\theta) = 0$ when $\theta - \alpha = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \ldots$ , that is when $\theta = \dfrac{\pi}{2} + \alpha, \dfrac{3\pi}{2} + \alpha, \ldots$ ;
$f(\theta) = 1$ when $\theta - \alpha = 0, 2\pi, 4\pi, \ldots$ , that is when $\theta = \alpha, 2\pi + \alpha, 4\pi + \alpha, \ldots$ ;
$f(\theta) = -1$ when $\theta - \alpha = \pi, 3\pi, 5\pi, \ldots$ , that is when $\theta = \pi + \alpha, 3\pi + \alpha, 5\pi + \alpha, \ldots$ .

So the graph of $f(\theta) = \cos\left(\theta - \dfrac{\pi}{4}\right)$ , for example, is identical in shape to the graph of $f(\theta) = \cos\theta$ , but is in a position given by moving the standard cosine curve a horizontal distance $\dfrac{\pi}{4}$ to the right. Similarly the graph of $f(\theta) = \cos(\theta + \alpha)$ is given by moving the standard cosine curve a horizontal distance $\alpha$ to the left, as the graph of $\cos\left(\theta + \dfrac{\pi}{3}\right)$ shows.

Figure 5.27. The graphs of

\cos\left(\theta - \frac{\pi}{4}\right)

and

\cos\left(\theta + \frac{\pi}{3}\right)

Now consider the function $f(\theta) = \sin(2\theta + \alpha)$ . Adding a constant angle moves a curve to the left, and in this case $\sin(2\theta + \alpha) = 0$ when $\theta = -\dfrac{\alpha}{2}$ , so the graph of this function is obtained by moving the graph of $\sin 2\theta$ a distance $\dfrac{\alpha}{2}$ to the left. For example, the graph of $\sin\left(2\theta + \dfrac{\pi}{2}\right)$ is that of $\sin 2\theta$ moved $\dfrac{\pi}{4}$ to the left.

Similarly, if $f(\theta) = \cos\left(3\theta - \dfrac{\pi}{4}\right)$ , putting $3\theta - \dfrac{\pi}{4} = 0$ shows that the graph is obtained by moving the graph of $\cos 3\theta$ a distance $\dfrac{\pi}{12}$ to the right. And if $f(\theta) = \tan\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right)$ , we have a tangent curve with a frequency of $\frac12$ which is moved a distance $\dfrac{\pi}{2}$ to the left.

Figure 5.28. The graphs of

\sin\left(2\theta + \frac{\pi}{2}\right)

\cos\left(3\theta - \frac{\pi}{4}\right)

and

\tan\left(\frac12\theta + \frac{\pi}{4}\right)

An equation containing a compound angle is solved in the same way as one containing a multiple angle.

Example 5.33.

Find the general solution of the equation $\cos\left(2\theta - \dfrac{\pi}{6}\right) = \dfrac12$ .

The principal value of $2\theta - \dfrac{\pi}{6}$ is $\dfrac{\pi}{3}$ , so

2\theta - \frac{\pi}{6} = \pm\frac{\pi}{3} + 2n\pi, \qquad 2\theta = \pm\frac{\pi}{3} + 2n\pi + \frac{\pi}{6}, \qquad \theta = \pm\frac{\pi}{6} + n\pi + \frac{\pi}{12},

that is $\theta = n\pi + \dfrac{\pi}{4}$ or $\theta = n\pi - \dfrac{\pi}{12}$ .

Problem 5.11.

Sketch the graphs of $\sin 4\theta$ and $\sec 2\theta$ in the domain $0 \leqslant \theta \leqslant 2\pi$ , and state the period and the frequency of each function.
Find the general solution of the equation $\cos\left(\theta - \dfrac{\pi}{6}\right) = -\dfrac12$ .

Trigonometric Identities

Any one angle has six trigonometric ratios, and a particular value of one ratio applies to an infinite set of angles, so it is not surprising that relationships exist between the various circular functions. They are identities, true for every angle for which the ratios are defined, and they are very useful in the development of trigonometry.

Consider first the relationship between the sine, cosine and tangent of any angle. With $P(x, y)$ as in Figure 5.29,

\sin\theta = \frac{y}{OP}, \qquad \cos\theta = \frac{x}{OP}, \qquad \tan\theta = \frac{y}{x} .

But $\dfrac{y}{x} = \dfrac{y}{OP} \div \dfrac{x}{OP}$ , so for all angles

\tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad\text{and similarly}\qquad \cot\theta = \frac{\cos\theta}{\sin\theta} .

Figure 5.29. The right-angled triangle

OPQ

for any position of

P

The Pythagorean Identities

For any position of $OP$ a right-angled triangle $OPQ$ can be drawn, for which, by Pythagoras,

x^2 + y^2 = OP^2 .

Dividing throughout, in turn, by $OP^2$ , $x^2$ and $y^2$ gives

\left(\frac{x}{OP}\right)^2 + \left(\frac{y}{OP}\right)^2 = 1, \qquad 1 + \left(\frac{y}{x}\right)^2 = \left(\frac{OP}{x}\right)^2, \qquad \left(\frac{x}{y}\right)^2 + 1 = \left(\frac{OP}{y}\right)^2 .

The use of brackets when raising a trigonometric ratio to a power can be avoided by writing $\cos^2\theta$ for $(\cos\theta)^2$ , and so on, and then these become

\cos^2\theta + \sin^2\theta = 1, \qquad 1 + \tan^2\theta = \sec^2\theta, \qquad \cot^2\theta + 1 = \operatorname{cosec}^2\theta .

These relationships are valid for any position of $OP$ , that is for all angles. They are very useful in the solution of certain trigonometric equations.

Example 5.34.

Solve the equation $2\cos^2\theta - \sin\theta = 1$ for values of $\theta$ between $0$ and $2\pi$ .

Using $\cos^2\theta + \sin^2\theta = 1$ gives

2(1 - \sin^2\theta) - \sin\theta = 1, \qquad\text{or}\qquad 2\sin^2\theta + \sin\theta - 1 = 0 .

This is now a quadratic equation in $\sin\theta$ , of the form $2x^2 + x - 1 = 0$ , hence

(2\sin\theta - 1)(\sin\theta + 1) = 0, \qquad \sin\theta = \tfrac12 \ \text{or}\ \sin\theta = -1 .

If $\sin\theta = \frac12$ , $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$ , and if $\sin\theta = -1$ , $\theta = \dfrac{3\pi}{2}$ . Therefore the solution of the equation is $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$ .

Example 5.35.

If $3\sec^2\theta - 5\tan\theta - 4 = 0$ , find the general solution.

Using $1 + \tan^2\theta = \sec^2\theta$ we have

3(1 + \tan^2\theta) - 5\tan\theta - 4 = 0, \qquad\text{that is}\qquad 3\tan^2\theta - 5\tan\theta - 1 = 0 .

Again we have a quadratic equation, but because it has no simple factors we solve it by the formula:

\tan\theta = \frac{5 \pm \sqrt{25 + 12}}{6} = 1.8471 \ \text{or}\ -0.1805 .

If $\tan\theta = 1.8471$ the principal solution is $\theta = 61.57^\circ$ , and if $\tan\theta = -0.1805$ it is $\theta = -10.23^\circ$ . The complete general solution is therefore

\theta = 180n^\circ + 61.57^\circ \qquad\text{or}\qquad \theta = 180n^\circ - 10.23^\circ .

Other applications of the standard identities include the derivation of further trigonometric relationships, the elimination of trigonometric terms from pairs of equations, and the calculation of the remaining trigonometric ratios of an angle for which only one ratio is known.

Example 5.36.

Show that $(1 - \cos A)(1 + \sec A) = \sin A\tan A$ .

Because this relationship has yet to be established, we must not assume that it is true by using the complete identity in our working: the left- and right-hand sides must be kept apart throughout. Considering the left-hand side,

\begin{aligned} (1 - \cos A)(1 + \sec A) &= 1 + \sec A - \cos A - \cos A\sec A = 1 + \sec A - \cos A - 1 \\ &= \frac{1}{\cos A} - \cos A = \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A} = \sin A\tan A, \end{aligned}

which is identical to the right-hand side.

Example 5.37.

Show that $(\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}$ .

Considering the left-hand side,

(\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right) = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} = \cos A\sin A .

This is already a very simple form, but it is not obviously identical to the right-hand side, so this time we work independently on the right-hand side:

\frac{1}{\tan A + \cot A} = 1 \div \left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right) = 1 \div \frac{\sin^2 A + \cos^2 A}{\cos A\sin A} = \cos A\sin A .

Since both sides reduce to $\cos A\sin A$ , they are identical.

Example 5.38.

Eliminate $\theta$ from the equations $x = 2\cos\theta$ and $y = 3\sin\theta$ .

Here $\cos\theta = \dfrac{x}{2}$ and $\sin\theta = \dfrac{y}{3}$ . Using $\cos^2\theta + \sin^2\theta = 1$ gives

\frac{x^2}{4} + \frac{y^2}{9} = 1, \qquad\text{that is}\qquad 9x^2 + 4y^2 = 36 .

In this problem both $x$ and $y$ depend on $\theta$ , a variable angle. Used in this way, $\theta$ is called a parameter.

Example 5.39.

If $\sin A = \frac13$ and $A$ is obtuse, find $\cos A$ and $\cot A$ without using a calculator.

Using $\cos^2 A + \sin^2 A = 1$ gives $\cos^2 A + \frac19 = 1$ , so $\cos^2 A = \frac89$ and $\cos A = \pm\dfrac{2\sqrt2}{3}$ . But $A$ is obtuse, so $\cos A$ is negative. Hence

\cos A = -\frac{2\sqrt2}{3}, \qquad \cot A = \frac{\cos A}{\sin A} = -2\sqrt2 .

This type of problem can often be done more directly by drawing the appropriate right-angled triangle and using Pythagoras.

Problem 5.12.

Solve the equation $\tan\theta + \cot\theta = 2$ for angles in the range $-180^\circ \leqslant \theta \leqslant 180^\circ$ .
Find the general solution of the equation $5\cos\theta - 4\sin^2\theta = 2$ .
Show that $\cot\theta + \tan\theta = \sec\theta\operatorname{cosec}\theta$ , and that $\dfrac{\sin A}{1 + \cos A} = \dfrac{1 - \cos A}{\sin A}$ .
Eliminate $\theta$ from the equations $x = 4\sec\theta$ and $y = 5\tan\theta$ .

Compound Angle Identities

It is often useful to be able to express the trigonometric ratios of angles such as $A + B$ or $A - B$ in terms of the ratios of $A$ and of $B$ . At first sight it is dangerously easy to think, for instance, that $\sin(A + B)$ is $\sin A + \sin B$ . That this is false can be seen by considering

\sin(45^\circ + 45^\circ) = \sin 90^\circ = 1, \qquad\text{whereas}\qquad \sin 45^\circ + \sin 45^\circ = \frac{\sqrt2}{2} + \frac{\sqrt2}{2} = \sqrt2 \neq 1 .

So the sine function is not distributive, and similarly for the other trigonometric ratios. The correct expression is

\sin(A + B) = \sin A\cos B + \cos A\sin B .

We show this geometrically when $A$ and $B$ are both acute, using Figure 5.30. The right-angled triangles $OPQ$ and $OQR$ contain the angles $A$ and $B$ , $RT$ is perpendicular to $OP$ , and $QS$ is perpendicular to $RT$ . Since $QS$ is parallel to $OP$ , the angle $SQO$ is $A$ , so the angle $SQR$ is $90^\circ - A$ and the angle $SRQ$ is equal to $A$ . Since $TS = PQ$ ,

\sin(A + B) = \frac{TR}{OR} = \frac{TS + SR}{OR} = \frac{PQ}{OR} + \frac{SR}{OR} = \frac{PQ}{OQ} \cdot \frac{OQ}{OR} + \frac{SR}{QR} \cdot \frac{QR}{OR} = \sin A\cos B + \cos A\sin B .

Figure 5.30. The construction for

\sin(A + B)

when

A

and

B

are acute.

Accepting at this stage that the formula is valid for all angles, it can be adapted to give the full set of compound angle identities.

Replacing $B$ by $-B$ , and using $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$ , gives $\sin(A - B) = \sin A\cos B - \cos A\sin B$ .
Replacing $A$ by $\dfrac{\pi}{2} - A$ in this identity, and using the complementary ratios, gives $\sin\left(\dfrac{\pi}{2} - (A + B)\right) = \cos A\cos B - \sin A\sin B$ , that is $\cos(A + B) = \cos A\cos B - \sin A\sin B$ .
Replacing $B$ by $-B$ in this identity gives $\cos(A - B) = \cos A\cos B + \sin A\sin B$ .
Dividing $\sin(A + B)$ by $\cos(A + B)$ , and then dividing the numerator and the denominator by $\cos A\cos B$ , gives $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$ .
Replacing $B$ by $-B$ in this identity, with $\tan(-B) = -\tan B$ , gives $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A\tan B}$ .

Collating these results we have

\begin{aligned} \sin(A + B) &= \sin A\cos B + \cos A\sin B, & \sin(A - B) &= \sin A\cos B - \cos A\sin B, \\ \cos(A + B) &= \cos A\cos B - \sin A\sin B, & \cos(A - B) &= \cos A\cos B + \sin A\sin B, \\ \tan(A + B) &= \frac{\tan A + \tan B}{1 - \tan A\tan B}, & \tan(A - B) &= \frac{\tan A - \tan B}{1 + \tan A\tan B} . \end{aligned}

The similarity between the pairs of identities for $A + B$ and $A - B$ makes it clear that care must be taken with signs when using these formulae.

Example 5.40.

Without using a calculator, evaluate $\sin 75^\circ$ , $\cos 105^\circ$ and $\tan(-15^\circ)$ .

\begin{aligned} \sin 75^\circ &= \sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{1}{\sqrt2} \cdot \frac{\sqrt3}{2} + \frac{1}{\sqrt2} \cdot \frac12 = \frac{\sqrt3 + 1}{2\sqrt2}, \\ \cos 105^\circ &= \cos(60^\circ + 45^\circ) = \cos 60^\circ\cos 45^\circ - \sin 60^\circ\sin 45^\circ = \frac12 \cdot \frac{1}{\sqrt2} - \frac{\sqrt3}{2} \cdot \frac{1}{\sqrt2} = \frac{1 - \sqrt3}{2\sqrt2}, \\ \tan(-15^\circ) &= \tan(45^\circ - 60^\circ) = \frac{\tan 45^\circ - \tan 60^\circ}{1 + \tan 45^\circ\tan 60^\circ} = \frac{1 - \sqrt3}{1 + \sqrt3} = \sqrt3 - 2, \end{aligned}

the last after rationalising the denominator. The value of $\cos 105^\circ$ is negative, which is consistent with the cosine of an angle in the second quadrant. In each part there are alternative compound angles which could be used, such as $75^\circ = 120^\circ - 45^\circ$ , $105^\circ = 150^\circ - 45^\circ$ and $-15^\circ = 30^\circ - 45^\circ$ .

Example 5.41.

$A$ is obtuse and $\sin A = \frac35$ , and $B$ is acute and $\sin B = \frac{12}{13}$ . Without finding the values of $A$ and $B$ , evaluate $\cos(A + B)$ and $\tan(A - B)$ .

In order to use the compound angle formulae we need $\cos A$ , $\cos B$ , $\tan A$ and $\tan B$ . These are most simply obtained by using Pythagoras in the appropriate right-angled triangles, with sides $3$ , $4$ , $5$ and $5$ , $12$ , $13$ , remembering that $A$ is obtuse:

\cos A = -\frac45, \quad \tan A = -\frac34, \qquad \cos B = \frac{5}{13}, \quad \tan B = \frac{12}{5} .

Then

\begin{aligned} \cos(A + B) &= \cos A\cos B - \sin A\sin B = \left(-\frac45\right)\left(\frac{5}{13}\right) - \left(\frac35\right)\left(\frac{12}{13}\right) = -\frac{56}{65}, \\ \tan(A - B) &= \frac{\tan A - \tan B}{1 + \tan A\tan B} = \frac{-\frac34 - \frac{12}{5}}{1 + \left(-\frac34\right)\left(\frac{12}{5}\right)} = \frac{63}{16} . \end{aligned}

Example 5.42.

Show that

\frac{\sin(A - B)}{\cos A\cos B} + \frac{\sin(B - C)}{\cos B\cos C} + \frac{\sin(C - A)}{\cos C\cos A} = 0 .

Expanding each numerator, the left-hand side becomes

\frac{\sin A\cos B - \cos A\sin B}{\cos A\cos B} + \frac{\sin B\cos C - \cos B\sin C}{\cos B\cos C} + \frac{\sin C\cos A - \cos C\sin A}{\cos C\cos A},

which is

(\tan A - \tan B) + (\tan B - \tan C) + (\tan C - \tan A) = 0 .

Example 5.43.

Solve the equation $2\cos\theta = \sin(\theta + 30^\circ)$ , giving the general values of $\theta$ .

It is very important to appreciate that this is not an identity: only certain values of $\theta$ satisfy the equation.

2\cos\theta = \sin\theta\cos 30^\circ + \cos\theta\sin 30^\circ = \frac{\sqrt3}{2}\sin\theta + \frac12\cos\theta,

therefore

\frac32\cos\theta = \frac{\sqrt3}{2}\sin\theta, \qquad\text{so}\qquad \tan\theta = \sqrt3 .

The principal solution is $\theta = 60^\circ$ , so the general solution is $\theta = 180n^\circ + 60^\circ$ .

Problem 5.13.

Without using a calculator, evaluate $\cos 80^\circ\cos 20^\circ + \sin 80^\circ\sin 20^\circ$ and $\sin 165^\circ$ .
Show that $\dfrac{\sin(A + B)}{\cos A\cos B} = \tan A + \tan B$ .
Solve the equation $\sin(x + 60^\circ) = \cos x$ , giving the angles from $0^\circ$ to $360^\circ$ .

Double Angle Identities

The compound angle formulae deal with any two angles $A$ and $B$ , so they can be used for two equal angles. Replacing $B$ by $A$ in the formulae for $A + B$ gives

\sin 2A = 2\sin A\cos A, \qquad \cos 2A = \cos^2 A - \sin^2 A, \qquad \tan 2A = \frac{2\tan A}{1 - \tan^2 A} .

The second of these can be expressed in several forms, because

\cos^2 A - \sin^2 A = (1 - \sin^2 A) - \sin^2 A = 1 - 2\sin^2 A, \qquad \cos^2 A - \sin^2 A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1 .

Thus

\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1,

and these alternative expressions can themselves be rearranged to give

2\sin^2 A = 1 - \cos 2A, \qquad 2\cos^2 A = 1 + \cos 2A .

Complete familiarity with all the double angle formulae, including all the alternative forms of $\cos 2A$ , is essential. They are probably the most useful of all the trigonometric identities for simplifying trigonometric functions.

Example 5.44.

Find the general solution of the equation $\cos 2x + 3\sin x = 2$ .

Using $\cos 2x = 1 - 2\sin^2 x$ gives

1 - 2\sin^2 x + 3\sin x = 2, \qquad 2\sin^2 x - 3\sin x + 1 = 0, \qquad (2\sin x - 1)(\sin x - 1) = 0,

so $\sin x = \frac12$ or $\sin x = 1$ .

For $\sin x = \frac12$ the principal solution is $x = \dfrac{\pi}{6}$ , and the general solution is $x = 2n\pi + \dfrac{\pi}{6}$ or $x = (2n + 1)\pi - \dfrac{\pi}{6}$ .
For $\sin x = 1$ the principal solution is $x = \dfrac{\pi}{2}$ , and the general solution is $x = 2n\pi + \dfrac{\pi}{2}$ .

The full solution is therefore $x = 2n\pi + \dfrac{\pi}{6}$ , $(2n + 1)\pi - \dfrac{\pi}{6}$ or $2n\pi + \dfrac{\pi}{2}$ .

Example 5.45.

If $\tan\theta = \frac34$ and $\theta$ is acute, find the values of $\tan 2\theta$ , $\tan 4\theta$ and $\tan\dfrac{\theta}{2}$ .

In this problem we use $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ for three cases.

\begin{aligned} A = \theta: &\qquad \tan 2\theta = \frac{2 \times \frac34}{1 - \frac{9}{16}} = \frac{24}{7}, \\ A = 2\theta: &\qquad \tan 4\theta = \frac{2 \times \frac{24}{7}}{1 - \frac{576}{49}} = -\frac{336}{527} . \end{aligned}

For $A = \dfrac{\theta}{2}$ , writing $t = \tan\dfrac{\theta}{2}$ ,

\frac{2t}{1 - t^2} = \frac34, \qquad 8t = 3 - 3t^2, \qquad 3t^2 + 8t - 3 = 0, \qquad (3t - 1)(t + 3) = 0,

so $t = \frac13$ or $t = -3$ . But $\theta$ is acute, so $\dfrac{\theta}{2}$ is acute and its tangent is positive. Therefore $\tan\dfrac{\theta}{2} = \dfrac13$ .

Example 5.46.

Show that $\sin 3A = 3\sin A - 4\sin^3 A$ .

\begin{aligned} \sin 3A = \sin(2A + A) &= \sin 2A\cos A + \cos 2A\sin A = 2\sin A\cos^2 A + (1 - 2\sin^2 A)\sin A \\ &= 2\sin A(1 - \sin^2 A) + \sin A - 2\sin^3 A = 3\sin A - 4\sin^3 A . \end{aligned}

This is quite a useful identity, and it is worth remembering.

Example 5.47.

Eliminate $\theta$ from the equations $x = \cos 2\theta$ and $y = \sec\theta$ .

Using $\cos 2\theta = 2\cos^2\theta - 1$ , and $\cos\theta = \dfrac{1}{y}$ ,

x = \frac{2}{y^2} - 1, \qquad\text{hence}\qquad (x + 1)y^2 = 2 .

This is a Cartesian equation, obtained by eliminating the parameter $\theta$ from a pair of parametric equations.

Problem 5.14.

Express as a single trigonometric ratio $2\sin 14^\circ\cos 14^\circ$ , $1 - 2\sin^2 40^\circ$ and $\dfrac{1 + \tan x}{1 - \tan x}$ .
Solve the equation $\cos 2x = \sin x$ for angles in the range $0^\circ \leqslant x \leqslant 360^\circ$ , and state the general solution.
Show that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ .

Identities and the Inverse Functions

The double angle and compound angle identities are often useful in simplifying expressions, or solving equations, which contain inverse trigonometric functions.

Example 5.48.

Find $x$ if $\arcsin x + \arccos\dfrac{x}{2} = \dfrac{5\pi}{6}$ .

Let $\arcsin x = \theta$ , so that $\sin\theta = x$ , and $\cos\theta = \sqrt{1 - x^2}$ since $-\dfrac{\pi}{2} \leqslant \theta \leqslant \dfrac{\pi}{2}$ . Let $\arccos\dfrac{x}{2} = \varphi$ , so that $\cos\varphi = \dfrac{x}{2}$ , and $\sin\varphi = \dfrac{\sqrt{4 - x^2}}{2}$ since $0 \leqslant \varphi \leqslant \pi$ . The given equation then becomes $\theta + \varphi = \dfrac{5\pi}{6}$ , so

\sin(\theta + \varphi) = \frac12, \qquad \sin\theta\cos\varphi + \cos\theta\sin\varphi = \frac12,

that is

\frac{x^2}{2} + \frac{\sqrt{1 - x^2}\,\sqrt{4 - x^2}}{2} = \frac12, \qquad \sqrt{1 - x^2}\,\sqrt{4 - x^2} = 1 - x^2 .

Squaring both sides, and not cancelling $\sqrt{1 - x^2}$ , which would lose solutions, gives

(1 - x^2)(4 - x^2) = (1 - x^2)^2, \qquad (1 - x^2)\bigl[(4 - x^2) - (1 - x^2)\bigr] = 3(1 - x^2) = 0,

so $x = \pm 1$ . But $\arcsin(-1) = -\dfrac{\pi}{2}$ and $\arccos\left(-\dfrac12\right) = \dfrac{2\pi}{3}$ , whose sum is $\dfrac{\pi}{6}$ , so $x = -1$ does not satisfy the given equation. The only solution is $x = 1$ .

Example 5.49.

Show that $\arctan 3 + 2\arctan 2 = \pi + \operatorname{arccot} 3$ .

Let $\theta = \arctan 3$ , so that $\tan\theta = 3$ and $\dfrac{\pi}{4} < \theta < \dfrac{\pi}{2}$ , and let $\varphi = \arctan 2$ , so that $\tan\varphi = 2$ and $\dfrac{\pi}{4} < \varphi < \dfrac{\pi}{2}$ . The left-hand side is then $\theta + 2\varphi$ . Now

\tan 2\varphi = \frac{2 \times 2}{1 - 4} = -\frac43, \qquad \tan(\theta + 2\varphi) = \frac{\tan\theta + \tan 2\varphi}{1 - \tan\theta\tan 2\varphi} = \frac{3 - \frac43}{1 - 3\left(-\frac43\right)} = \frac{\frac53}{5} = \frac13 .

But $\dfrac{\pi}{4} < \theta < \dfrac{\pi}{2}$ and $\dfrac{\pi}{4} < \varphi < \dfrac{\pi}{2}$ give $\dfrac{3\pi}{4} < \theta + 2\varphi < \dfrac{3\pi}{2}$ , and the only angle in this range with tangent $\frac13$ is $\pi + \arctan\frac13$ . So

\arctan 3 + 2\arctan 2 = \pi + \arctan\frac13 = \pi + \operatorname{arccot} 3 .

Example 5.50.

Simplify $\arctan x + \arctan\dfrac{1 - x}{1 + x}$ , where $x > -1$ .

Let $\alpha = \arctan x$ and $\beta = \arctan\dfrac{1 - x}{1 + x}$ , so that $\tan\alpha = x$ and $\tan\beta = \dfrac{1 - x}{1 + x}$ , and we have to simplify $\alpha + \beta$ . Using $\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$ gives

\tan(\alpha + \beta) = \frac{x + \dfrac{1 - x}{1 + x}}{1 - x \cdot \dfrac{1 - x}{1 + x}} = \frac{x + x^2 + 1 - x}{1 + x - x + x^2} = 1 .

Since $x > -1$ , both $x$ and $\dfrac{1 - x}{1 + x}$ are greater than $-1$ , so $\alpha$ and $\beta$ each lie between $-\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$ , and $\alpha + \beta$ lies between $-\dfrac{\pi}{2}$ and $\pi$ . The only angle in that range whose tangent is $1$ is $\dfrac{\pi}{4}$ . Thus

\arctan x + \arctan\frac{1 - x}{1 + x} = \frac{\pi}{4} .

Problem 5.15.

Show that $\arctan\dfrac13 + \arctan\dfrac12 = \dfrac{\pi}{4}$ .
Solve the equation $\arctan(1 + x) + \arctan(1 - x) = \arctan 2$ .
Simplify $\sin(2\arctan x)$ .

The Half Angle Identities

We already know that $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ . Dividing $\sin 2A = 2\sin A\cos A$ and $\cos 2A = \cos^2 A - \sin^2 A$ by $\cos^2 A + \sin^2 A$ , which is $1$ , and then dividing the numerator and the denominator by $\cos^2 A$ , gives also

\sin 2A = \frac{2\tan A}{1 + \tan^2 A}, \qquad \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} .

If we replace $2A$ by $\theta$ and use $t$ to denote $\tan\dfrac{\theta}{2}$ , we have

\tan\theta = \frac{2t}{1 - t^2}, \qquad \sin\theta = \frac{2t}{1 + t^2}, \qquad \cos\theta = \frac{1 - t^2}{1 + t^2} .

These three identities allow all the trigonometric ratios of any one angle to be expressed in terms of a common variable $t$ . In problems where none of the identities used so far can be applied, this group can be helpful.

Example 5.51.

Solve the equation $\sin\theta + 2\cos\theta = 1$ for angles between $0^\circ$ and $360^\circ$ .

With $t = \tan\dfrac{\theta}{2}$ ,

\frac{2t}{1 + t^2} + \frac{2(1 - t^2)}{1 + t^2} = 1, \qquad 2t + 2 - 2t^2 = 1 + t^2, \qquad 3t^2 - 2t - 1 = 0, \qquad (3t + 1)(t - 1) = 0 .

Therefore either $\tan\dfrac{\theta}{2} = -\dfrac13$ or $\tan\dfrac{\theta}{2} = 1$ . The range of values specified for $\theta$ is $0^\circ$ to $360^\circ$ , so the range of values required for $\dfrac{\theta}{2}$ is $0^\circ$ to $180^\circ$ . Within this range $\tan\dfrac{\theta}{2} = -\dfrac13$ gives $\dfrac{\theta}{2} = 161.57^\circ$ , and $\tan\dfrac{\theta}{2} = 1$ gives $\dfrac{\theta}{2} = 45^\circ$ . Thus $\theta = 323.13^\circ, 90^\circ$ .

Extra care is sometimes needed with this method, as a very similar equation shows. Consider

\sin\theta - \cos\theta = 1 .

Using $t = \tan\dfrac{\theta}{2}$ gives

\frac{2t}{1 + t^2} - \frac{1 - t^2}{1 + t^2} = 1, \qquad 2t - 1 + t^2 = 1 + t^2 .

The two $t^2$ terms cancel, leaving $2t = 2$ , so $t = 1$ . But $t = \tan\dfrac{\theta}{2}$ is not defined when $\dfrac{\theta}{2}$ is an odd multiple of $\dfrac{\pi}{2}$ , that is when $\theta = (2n + 1)\pi$ , and these angles have to be checked separately: $\sin\pi - \cos\pi = 0 + 1 = 1$ , so they are solutions as well. Hence $\tan\dfrac{\theta}{2} = 1$ , or $\tan\dfrac{\theta}{2}$ is undefined, giving

\frac{\theta}{2} = n\pi + \frac{\pi}{4} \quad\text{or}\quad \frac{\theta}{2} = n\pi + \frac{\pi}{2}, \qquad\text{that is}\qquad \theta = 2n\pi + \frac{\pi}{2} \quad\text{or}\quad \theta = (2n + 1)\pi .

Remark.

When the half angle identities are used, $t$ does not always represent $\tan\dfrac{\theta}{2}$ . For instance, in solving the equation $\sin 4\theta + \tan 2\theta = 0$ we would use $t = \tan 2\theta$ .

The Expression $a\cos\theta + b\sin\theta$

It is often useful to reduce $a\cos\theta + b\sin\theta$ to a single term such as $r\cos(\theta - \alpha)$ . This is possible provided that we can find values of $r$ and $\alpha$ for which

r(\cos\theta\cos\alpha + \sin\theta\sin\alpha) = a\cos\theta + b\sin\theta

for every $\theta$ . Comparing the coefficients of $\cos\theta$ and of $\sin\theta$ ,

r\cos\alpha = a, \qquad r\sin\alpha = b .

Squaring and adding gives $r^2 = a^2 + b^2$ , so $r$ is equal to the length of the hypotenuse of the triangle containing $\alpha$ with sides $a$ and $b$ , and dividing gives $\tan\alpha = \dfrac{b}{a}$ , with $\alpha$ in the quadrant given by the signs of $a$ and $b$ . Thus

a\cos\theta + b\sin\theta = r\cos(\theta - \alpha), \qquad\text{where}\qquad r = \sqrt{a^2 + b^2} \quad\text{and}\quad \tan\alpha = \frac{b}{a} .

Example 5.52.

Express $3\cos\theta + 4\sin\theta$ in the form $r\cos(\theta - \alpha)$ , giving the values of $r$ and $\alpha$ .

Let

r(\cos\theta\cos\alpha + \sin\theta\sin\alpha) = 3\cos\theta + 4\sin\theta,

so that $r\cos\alpha = 3$ and $r\sin\alpha = 4$ . Thus $r = 5$ and $\tan\alpha = \dfrac43$ , which gives $\alpha = 53.13^\circ$ , and

3\cos\theta + 4\sin\theta = 5\cos(\theta - 53.13^\circ) .

It is sometimes more convenient to begin by comparing $a\cos\theta + b\sin\theta$ with $r\sin(\theta + \alpha)$ , so that

r(\sin\theta\cos\alpha + \cos\theta\sin\alpha) = a\cos\theta + b\sin\theta, \qquad r\sin\alpha = a, \quad r\cos\alpha = b .

Then $\tan\alpha = \dfrac{a}{b}$ and $r = \sqrt{a^2 + b^2}$ . The value of $\alpha$ is not the same as it was when we used $r\cos(\theta - \alpha)$ . Further variations that could be used are $r\sin(\theta - \alpha)$ and $r\cos(\theta + \alpha)$ . When using this method it is better to work from the basic comparison each time, as in the example above, than to quote values of $r$ and $\alpha$ .

Problem 5.16.

Express $5\cos\theta + 12\sin\theta$ in the forms $r\cos(\theta - \alpha)$ and $r\sin(\theta + \alpha)$ .
Express $3\sin\theta + 4\cos\theta$ in the forms $r\sin(\theta + \alpha)$ and $r\cos(\theta - \alpha)$ .
Express $\sqrt3\cos\theta - \sin\theta$ in the form $r\cos(\theta + \alpha)$ .

The Graph of $a\cos\theta + b\sin\theta$

First consider the function $f(\theta) = k\cos\theta$ , where $k > 0$ , which has the following characteristics.

$f(\theta) = 0$ when $\theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \ldots$ .
$f(\theta) = k$ when $\theta = 0, 2\pi, 4\pi, \ldots$ .
$f(\theta) = -k$ when $\theta = \pi, 3\pi, 5\pi, \ldots$ .

So the graph of this function is very similar to a standard cosine curve, but has maximum and minimum values $\pm k$ ; we say that the curve has an amplitude of $k$ . We also saw that the graph of $\cos(\theta - \alpha)$ is given by moving the graph of $\cos\theta$ a distance $\alpha$ to the right. Combining these two modifications of a standard cosine curve, the graph of the function $f(\theta) = k\cos(\theta - \alpha)$ can be sketched.

Figure 5.31. A cosine curve with amplitude

5

, and a cosine curve with amplitude

k

moved a distance

\alpha

to the right.

Consider now the function $a\cos\theta + b\sin\theta$ . At first sight its graph is not easy to visualise, but using $a\cos\theta + b\sin\theta = r\cos(\theta - \alpha)$ we see that its graph is a cosine curve modified in two ways.

Its maximum and minimum values are $\pm r$ , that is its amplitude is $r$ .
Its position is a distance $\alpha$ to the right of the standard curve.

Example 5.53.

Sketch the graph of the function $3\cos\theta + 4\sin\theta$ from $-180^\circ$ to $180^\circ$ .

As in the last example, $3\cos\theta + 4\sin\theta = r\cos(\theta - \alpha)$ with $r = 5$ and $\tan\alpha = \dfrac43$ , so $\alpha = 53.13^\circ$ . Hence the graph is a cosine curve with an amplitude of $5$ and a phase shift of $53.13^\circ$ to the right.

Figure 5.32. The graph of

3\cos\theta + 4\sin\theta

, with

\cos\theta

dashed.

It is interesting to see how the same graph is produced if the alternative form $3\cos\theta + 4\sin\theta = r\sin(\theta + \alpha')$ is used. With this approach $r = 5$ and $\tan\alpha' = \dfrac34$ , so that $3\cos\theta + 4\sin\theta = 5\sin(\theta + \alpha')$ , and the graph is a sine curve with an amplitude of $5$ and a displacement of $\alpha'$ to the left. But since $\tan\alpha' = \cot\alpha$ , the angles $\alpha'$ and $\alpha$ are complementary, that is $\alpha + \alpha' = \dfrac{\pi}{2}$ . We also know that a cosine curve is the same as a sine curve displaced $\dfrac{\pi}{2}$ to the left. So a cosine curve moved a distance $\alpha$ to the right coincides with a sine curve moved a distance $\alpha'$ to the left.

So any correct compound angle form of $a\cos\theta + b\sin\theta$ gives a quick method of sketching the graph of that function, and in particular of finding its maximum and minimum values, which for sine and cosine functions are also the greatest and least values.

The Equation $a\cos\theta + b\sin\theta = c$

One way of solving an equation of this type, using the half angle formulae, has already been used. A compound angle form gives an alternative method. Applied to the equation $\sin\theta + 2\cos\theta = 1$ , solved above with $t$ , it gives

2\cos\theta + \sin\theta = r(\cos\theta\cos\alpha + \sin\theta\sin\alpha) = r\cos(\theta - \alpha),

where $r\cos\alpha = 2$ and $r\sin\alpha = 1$ , that is $\tan\alpha = \frac12$ and $r = \sqrt5$ . Hence

\sqrt5\cos(\theta - \alpha) = 1, \qquad \cos(\theta - \alpha) = \frac{1}{\sqrt5}, \qquad \theta - \alpha = 360n^\circ \pm 63.43^\circ,

from which $\theta = 360n^\circ \pm 63.43^\circ + \alpha$ . But $\alpha = \arctan\frac12 = 26.57^\circ$ , so $\theta = 360n^\circ + 90^\circ$ or $\theta = 360n^\circ - 36.87^\circ$ . Using the values of $n$ which give $\theta$ between $0^\circ$ and $360^\circ$ , that is $n = 0$ and $n = 1$ , we have $\theta = 90^\circ, 323.13^\circ$ .

Example 5.54.

Express $\sqrt{\dfrac{1 - \sin 2\theta}{1 + \sin 2\theta}}$ in terms of $\tan\theta$ .

Using $\sin 2\theta = \dfrac{2t}{1 + t^2}$ , where $t = \tan\theta$ , gives

1 - \sin 2\theta = \frac{1 + t^2 - 2t}{1 + t^2} = \frac{(1 - t)^2}{1 + t^2}, \qquad 1 + \sin 2\theta = \frac{1 + t^2 + 2t}{1 + t^2} = \frac{(1 + t)^2}{1 + t^2} .

Hence

\frac{1 - \sin 2\theta}{1 + \sin 2\theta} = \frac{(1 - t)^2}{(1 + t)^2}, \qquad\text{so}\qquad \sqrt{\frac{1 - \sin 2\theta}{1 + \sin 2\theta}} = \pm\frac{1 - \tan\theta}{1 + \tan\theta},

taking the sign which makes the right-hand side positive.

Example 5.55.

Find the general solution of the equation $\cos\theta - \sqrt3\sin\theta = 1$ , first by using the half angle formulae and then by using a compound angle form.

With $t = \tan\dfrac{\theta}{2}$ , the equation becomes $\dfrac{1 - t^2 - 2\sqrt3\,t}{1 + t^2} = 1$ , so $1 - t^2 - 2\sqrt3\,t = 1 + t^2$ , that is $2t(t + \sqrt3) = 0$ . Hence either $t = 0$ or $t = -\sqrt3$ , and the principal values of $\dfrac{\theta}{2}$ are $0$ and $-\dfrac{\pi}{3}$ . So $\dfrac{\theta}{2} = n\pi$ or $\dfrac{\theta}{2} = n\pi - \dfrac{\pi}{3}$ , and $\theta = 2n\pi$ or $\theta = 2n\pi - \dfrac{2\pi}{3}$ . When $t$ is undefined, $\theta = (2n + 1)\pi$ , the left-hand side is $-1$ , so no solutions are lost.
Let $\cos\theta - \sqrt3\sin\theta = r(\cos\theta\cos\alpha - \sin\theta\sin\alpha) = r\cos(\theta + \alpha)$ , where $r\cos\alpha = 1$ and $r\sin\alpha = \sqrt3$ . Then $\tan\alpha = \sqrt3$ , so $\alpha = \dfrac{\pi}{3}$ and $r = 2$ , and the equation can be written $2\cos\left(\theta + \dfrac{\pi}{3}\right) = 1$ . The principal value of $\theta + \dfrac{\pi}{3}$ is $\dfrac{\pi}{3}$ , so $\theta + \dfrac{\pi}{3} = 2n\pi \pm \dfrac{\pi}{3}$ , and again $\theta = 2n\pi$ or $\theta = 2n\pi - \dfrac{2\pi}{3}$ .

Example 5.56.

Express $5\sin\theta + 12\cos\theta$ in the form $r\sin(\theta + \alpha)$ , giving the values of $r$ and $\alpha$ . Show that $5\sin\theta + 12\cos\theta + 7 \leqslant 20$ , and find the minimum value of $5\sin\theta + 12\cos\theta + 7$ . Sketch the graph of the function $\dfrac{1}{5\sin\theta + 12\cos\theta}$ for $0 \leqslant \theta \leqslant 2\pi$ .

Let $5\sin\theta + 12\cos\theta = r(\sin\theta\cos\alpha + \cos\theta\sin\alpha) = r\sin(\theta + \alpha)$ , so that $r\cos\alpha = 5$ and $r\sin\alpha = 12$ , giving $r = 13$ and $\tan\alpha = \dfrac{12}{5}$ , $\alpha = 67.38^\circ$ . Hence

5\sin\theta + 12\cos\theta = 13\sin(\theta + \alpha) .

But $-1 \leqslant \sin(\theta + \alpha) \leqslant 1$ , so $-13 \leqslant 5\sin\theta + 12\cos\theta \leqslant 13$ , and adding $7$ throughout gives

-6 \leqslant 5\sin\theta + 12\cos\theta + 7 \leqslant 20 .

This shows that $5\sin\theta + 12\cos\theta + 7 \leqslant 20$ , and that the minimum value of $5\sin\theta + 12\cos\theta + 7$ is $-6$ .

Now $5\sin\theta + 12\cos\theta = 13\sin(\theta + \alpha)$ , so the graph of $\dfrac{1}{5\sin\theta + 12\cos\theta}$ is also the graph of $\dfrac{1}{13}\operatorname{cosec}(\theta + \alpha)$ . Its general shape is a typical cosecant curve, except that its branches turn at the values $\dfrac{1}{13}$ and $-\dfrac{1}{13}$ , and its position is a distance $\alpha$ to the left of the standard curve.

Figure 5.33. The graph of

\dfrac{1}{5\sin\theta + 12\cos\theta}

for

0 \leqslant \theta \leqslant 2\pi

Problem 5.17.

Using $t = \tan\dfrac{\theta}{2}$ , solve the equation $3\cos\theta + 2\sin\theta = 3$ , giving the values of $\theta$ from $-180^\circ$ to $180^\circ$ .
Find the maximum and minimum values of $7\cos\theta - 24\sin\theta + 3$ , and the values of $\theta$ between $0^\circ$ and $360^\circ$ at which they occur.
Find the general solution of the equation $\cos x + \sin x = \sqrt2$ .

The Factor Formulae

To factorise is to express in the form of a product. The set of identities called the factor formulae converts expressions such as $\sin A + \sin B$ into a product. To derive them we use the compound angle group. Adding and subtracting

\sin A\cos B + \cos A\sin B = \sin(A + B), \qquad \sin A\cos B - \cos A\sin B = \sin(A - B)

gives the first two of the following identities, and similar treatment of $\cos(A + B)$ and $\cos(A - B)$ gives the other two:

\begin{aligned} 2\sin A\cos B &= \sin(A + B) + \sin(A - B), &\qquad &(1) \\ 2\cos A\sin B &= \sin(A + B) - \sin(A - B), & &(2) \\ 2\cos A\cos B &= \cos(A + B) + \cos(A - B), & &(3) \\ -2\sin A\sin B &= \cos(A + B) - \cos(A - B). & &(4) \end{aligned}

Putting $A + B = P$ and $A - B = Q$ , so that $A = \dfrac{P + Q}{2}$ and $B = \dfrac{P - Q}{2}$ , these become

\begin{aligned} \sin P + \sin Q &= 2\sin\frac{P + Q}{2}\cos\frac{P - Q}{2}, &\qquad &(5) \\ \sin P - \sin Q &= 2\cos\frac{P + Q}{2}\sin\frac{P - Q}{2}, & &(6) \\ \cos P + \cos Q &= 2\cos\frac{P + Q}{2}\cos\frac{P - Q}{2}, & &(7) \\ \cos P - \cos Q &= -2\sin\frac{P + Q}{2}\sin\frac{P - Q}{2}. & &(8) \end{aligned}

Identities $(5)$ to $(8)$ are best used when a sum or difference is to be expressed as a product, while identities $(1)$ to $(4)$ should be used when a given product is to be changed into a sum or difference. For example, to express $\sin 6\theta - \sin 4\theta$ as a product we use $(6)$ :

\sin 6\theta - \sin 4\theta = 2\cos\frac{6\theta + 4\theta}{2}\sin\frac{6\theta - 4\theta}{2} = 2\cos 5\theta\sin\theta .

But to express $2\cos 7\theta\cos 2\theta$ as a sum we use $(3)$ :

2\cos 7\theta\cos 2\theta = \cos(7\theta + 2\theta) + \cos(7\theta - 2\theta) = \cos 9\theta + \cos 5\theta .

When these identities are used regularly they are not too difficult to remember. Most people find it best to memorise them in words rather than as symbols. For example, $(5)$ can be remembered as “the sum of two sines is twice the sine of the semi-sum times the cosine of the semi-difference”, and $(1)$ as “twice sin cos is the sine of the sum plus the sine of the difference”. Identities $(4)$ and $(8)$ need special care because of the minus sign.

Example 5.57.

Show that $\dfrac{\sin A + \sin B}{\cos A + \cos B} = \tan\dfrac{A + B}{2}$ . If $A$ , $B$ and $C$ are the angles of a triangle, deduce that $\dfrac{\sin A + \sin B}{\cos A + \cos B} = \cot\dfrac{C}{2}$ .

Considering the left-hand side, by $(5)$ and $(7)$ ,

\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}\cos\frac{A - B}{2}} = \tan\frac{A + B}{2} .

If $A$ , $B$ and $C$ are the angles of a triangle, $A + B + C = 180^\circ$ , so $\dfrac{A + B}{2} + \dfrac{C}{2} = 90^\circ$ . The angles $\dfrac{A + B}{2}$ and $\dfrac{C}{2}$ are complementary, therefore $\tan\dfrac{A + B}{2} = \cot\dfrac{C}{2}$ , and

\frac{\sin A + \sin B}{\cos A + \cos B} = \cot\frac{C}{2} .

Example 5.58.

Solve the equation $\sin 5x - \sin 3x = 0$ , giving the general solution.

By $(6)$ ,

\sin 5x - \sin 3x = 2\cos\frac{5x + 3x}{2}\sin\frac{5x - 3x}{2} = 2\cos 4x\sin x = 0,

so either $\cos 4x = 0$ or $\sin x = 0$ . If $\cos 4x = 0$ , then $4x = (2n + 1)\dfrac{\pi}{2}$ , and if $\sin x = 0$ , then $x = n\pi$ . The general solution is therefore

x = (2n + 1)\frac{\pi}{8} \qquad\text{or}\qquad x = n\pi .

Example 5.59.

Factorise $\cos\theta - \cos 3\theta - \cos 5\theta + \cos 7\theta$ .

Grouping the terms in pairs,

(\cos 7\theta + \cos\theta) - (\cos 5\theta + \cos 3\theta) = 2\cos 4\theta\cos 3\theta - 2\cos 4\theta\cos\theta = 2\cos 4\theta\,(\cos 3\theta - \cos\theta),

and by $(8)$ , $\cos 3\theta - \cos\theta = -2\sin 2\theta\sin\theta$ . So

\cos\theta - \cos 3\theta - \cos 5\theta + \cos 7\theta = -4\cos 4\theta\sin 2\theta\sin\theta .

Other groupings of the four terms can be used, but arranging them so that pairs of cosines are added makes the factorising simplest.

Example 5.60.

If $A$ , $B$ and $C$ are the angles of a triangle, show that $\sin A + \sin B + \sin C = 4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$ .

Now $A + B + C = 180^\circ$ , so $\dfrac{A + B}{2} = 90^\circ - \dfrac{C}{2}$ , and therefore $\sin\dfrac{A + B}{2} = \cos\dfrac{C}{2}$ and $\sin\dfrac{C}{2} = \cos\dfrac{A + B}{2}$ . Considering the left-hand side, and using $(5)$ and the double angle formula for $\sin C$ ,

\begin{aligned} \sin A + \sin B + \sin C &= 2\sin\frac{A + B}{2}\cos\frac{A - B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2} \\ &= 2\cos\frac{C}{2}\cos\frac{A - B}{2} + 2\cos\frac{A + B}{2}\cos\frac{C}{2} \\ &= 2\cos\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\frac{A + B}{2}\right] . \end{aligned}

But by $(7)$ , $\cos\dfrac{A - B}{2} + \cos\dfrac{A + B}{2} = 2\cos\dfrac{A}{2}\cos\dfrac{B}{2}$ , so

\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} .

Problem 5.18.

Factorise $\sin 3A + \sin A$ and $\cos 7A - \cos A$ .
Solve the equation $\cos 2x + \cos 4x = 0$ , giving the values of $x$ from $0^\circ$ to $360^\circ$ .
If $A$ , $B$ and $C$ are the angles of a triangle, show that $\cos(B + C) = -\cos A$ and that $\cos A + \cos B + \cos C = 1 + 4\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$ .

Small Angles

A glance at the values of $\sin\theta$ and $\tan\theta$ when $\theta$ is a very small positive angle shows that these two ratios are almost equal. Further, if the small angle is measured in radians, both are found to be almost equal to $\theta$ . These relationships can be demonstrated as follows.

Consider a small angle $\theta$ , measured in radians, subtended by an arc $AB$ at the centre $O$ of a circle of radius $r$ . The area of the sector $OAB$ is $\frac12 r^2\theta$ . If $AC$ is drawn perpendicular to $OA$ , to cut $OB$ produced at $C$ , then $OAC$ is a right-angled triangle with base $r$ and height $r\tan\theta$ , so its area is $\frac12 r^2\tan\theta$ . Further, when the chord $AB$ is drawn, an isosceles triangle $OAB$ is formed with base $r$ and height $r\sin\theta$ , so its area is $\frac12 r^2\sin\theta$ .

Figure 5.34. The triangle

OAB

, the sector

OAB

and the triangle

OAC

Now area of triangle $OAB <$ area of sector $OAB <$ area of triangle $OAC$ , that is

\tfrac12 r^2\sin\theta < \tfrac12 r^2\theta < \tfrac12 r^2\tan\theta .

Dividing throughout by $\frac12 r^2$ , which is positive, gives $\sin\theta < \theta < \tan\theta$ . But $\sin\theta$ , $\theta$ and $\tan\theta$ are all positive, since $\theta$ is a small positive angle, so we can divide throughout by any of them. Dividing by $\sin\theta$ ,

1 < \frac{\theta}{\sin\theta} < \sec\theta .

For small values of $\theta$ , $\sec\theta \to 1$ as $\theta \to 0$ . Hence as $\theta \to 0$ , $\dfrac{\theta}{\sin\theta}$ lies between $1$ and a number which approaches $1$ , and we can say that $\dfrac{\theta}{\sin\theta} \to 1$ as $\theta \to 0$ , a limit in the sense of the last lesson. Similarly, by dividing the first inequalities by $\tan\theta$ , we get $\cos\theta < \dfrac{\theta}{\tan\theta} < 1$ , which shows that $\dfrac{\theta}{\tan\theta} \to 1$ as $\theta \to 0$ .

The same results are obtained if $\theta$ is a small negative angle, since $\dfrac{\theta}{\sin\theta}$ and $\dfrac{\theta}{\tan\theta}$ are unchanged when $\theta$ is replaced by $-\theta$ . These limiting values show that, for small values of $\theta$ ,

\sin\theta \approx \theta \qquad\text{and}\qquad \tan\theta \approx \theta .

So far we have not found an approximate value for $\cos\theta$ when $\theta$ is small. To do this we use the double angle identity

\cos\theta = 1 - 2\sin^2\frac{\theta}{2} .

If $\theta$ is small then so is $\dfrac{\theta}{2}$ , and $\sin\dfrac{\theta}{2} \approx \dfrac{\theta}{2}$ , so $\cos\theta \approx 1 - 2\left(\dfrac{\theta}{2}\right)^2 = 1 - \dfrac{\theta^2}{2}$ .

Thus, when $\theta$ is measured in radians,

\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1 \qquad\text{and}\qquad \lim_{\theta \to 0}\frac{\tan\theta}{\theta} = 1,

and for any small angle $\theta$ , measured in radians,

\sin\theta \approx \theta, \qquad \tan\theta \approx \theta, \qquad \cos\theta \approx 1 - \frac{\theta^2}{2} .

These approximations are correct to three significant figures for angles in the range $-0.105 \leqslant \theta \leqslant 0.105$ , that is about $-6^\circ \leqslant \theta \leqslant 6^\circ$ .

Example 5.61.

Find an approximation for the expression $\dfrac{\sin 3\theta}{1 + \cos 2\theta}$ when $\theta$ is small.

When $\theta$ is small, $3\theta$ is small, so $\sin 3\theta \approx 3\theta$ ; also $2\theta$ is small, so $\cos 2\theta \approx 1 - \dfrac{(2\theta)^2}{2} = 1 - 2\theta^2$ . So when $\theta$ is small,

\frac{\sin 3\theta}{1 + \cos 2\theta} \approx \frac{3\theta}{2 - 2\theta^2} .

Example 5.62.

Without using a calculator, find an approximate value for $\tan 61^\circ$ , given that $\sqrt3 = 1.732$ and $1^\circ = 0.017$ radians, giving the answer to three decimal places.

\tan 61^\circ = \tan(60^\circ + 1^\circ) = \frac{\tan 60^\circ + \tan 1^\circ}{1 - \tan 60^\circ\tan 1^\circ} = \frac{\sqrt3 + \tan 1^\circ}{1 - \sqrt3\tan 1^\circ} .

Now $\tan\theta \approx \theta$ when $\theta$ is small and measured in radians, so $\tan 1^\circ \approx 0.017$ . Hence

\tan 61^\circ \approx \frac{1.732 + 0.017}{1 - (1.732)(0.017)} = 1.802 .

Problem 5.19.

If $\theta$ is small, find approximations for $\dfrac{\theta\sin\theta}{1 - \cos\theta}$ and $\dfrac{\sin 4\theta}{\theta}$ .
If $\theta$ is small enough for $\theta^2$ to be neglected, show that $\tan\left(\dfrac{\pi}{4} + \theta\right) \approx \dfrac{1 + \theta}{1 - \theta}$ .

Further Properties of Triangles

There is a great variety of relationships between the sides and angles of a triangle besides the sine and cosine rules, and some of the most useful can be derived from the identities above.

The Cotangent Formula

If $D$ divides the side $AB$ of a triangle $ABC$ in the ratio $m : n$ then, with the angles marked in Figure 5.35,

(m + n)\cot\theta = m\cot\alpha - n\cot\beta .

This relationship is known as the cotangent formula, and we show it by using the sine rule in the triangles $ACD$ and $BCD$ . In the triangle $ACD$ the angle at $D$ is $180^\circ - \theta$ , so the angle $A$ is $\theta - \alpha$ , and

\frac{CD}{\sin A} = \frac{AD}{\sin\alpha}, \qquad CD = \frac{AD\sin(\theta - \alpha)}{\sin\alpha} .

In the triangle $BCD$ the angle $B$ is $180^\circ - \theta - \beta$ , so $\sin B = \sin(\theta + \beta)$ , and

\frac{CD}{\sin B} = \frac{BD}{\sin\beta}, \qquad CD = \frac{BD\sin(\theta + \beta)}{\sin\beta} .

Hence

\frac{AD\sin(\theta - \alpha)}{\sin\alpha} = \frac{BD\sin(\theta + \beta)}{\sin\beta} .

But $AD : DB = m : n$ , so

\frac{m\sin(\theta - \alpha)}{\sin\alpha} = \frac{n\sin(\theta + \beta)}{\sin\beta}, \qquad\text{that is}\qquad \frac{m(\sin\theta\cos\alpha - \cos\theta\sin\alpha)}{\sin\alpha} = \frac{n(\sin\theta\cos\beta + \cos\theta\sin\beta)}{\sin\beta} .

Dividing both sides by $\sin\theta$ gives

m(\cot\alpha - \cot\theta) = n(\cot\beta + \cot\theta), \qquad\text{so}\qquad (m + n)\cot\theta = m\cot\alpha - n\cot\beta .

If $D$ is the midpoint of $AB$ , this becomes $2\cot\theta = \cot\alpha - \cot\beta$ .

Figure 5.35. The angles in the cotangent formula, with

AD : DB = m : n

Other Relationships

The Projection Formula

In any triangle $ABC$ ,

c = b\cos A + a\cos B .

When $A$ and $B$ are acute, the perpendicular from $C$ to $AB$ divides $AB$ into two parts of lengths $b\cos A$ and $a\cos B$ , as in Figure 5.36. If one of the angles is obtuse, say $B$ , the foot of the perpendicular lies beyond $B$ , and the part $a\cos B$ is negative, so the formula still holds.

The Difference of Two Sides

In any triangle $ABC$ ,

\frac{a - b}{a + b} = \tan\frac{A - B}{2}\tan\frac{C}{2} .

We show this by using the sine rule in the form $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = k$ , say, so that $a = k\sin A$ and $b = k\sin B$ . Hence, by the factor formulae,

\frac{a - b}{a + b} = \frac{k(\sin A - \sin B)}{k(\sin A + \sin B)} = \frac{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}} = \tan\frac{A - B}{2}\cot\frac{A + B}{2} .

But $A + B + C = \pi$ , so $\dfrac{A + B}{2} = \dfrac{\pi}{2} - \dfrac{C}{2}$ , and $\cot\dfrac{A + B}{2} = \tan\dfrac{C}{2}$ . So

\frac{a - b}{a + b} = \tan\frac{A - B}{2}\tan\frac{C}{2} .

The Bisector of an Angle

In a triangle $ABC$ the bisector of the angle $A$ divides $BC$ in the ratio $c : b$ .

We show this by letting the bisector meet $BC$ at $D$ , with the angle $ADB$ equal to $\theta$ . In the triangle $ABD$ ,

\frac{BD}{\sin\frac12 A} = \frac{c}{\sin\theta},

and in the triangle $ACD$ , where the angle at $D$ is $180^\circ - \theta$ ,

\frac{DC}{\sin\frac12 A} = \frac{b}{\sin(180^\circ - \theta)} = \frac{b}{\sin\theta} .

Hence

\frac{\sin\frac12 A}{\sin\theta} = \frac{BD}{c} = \frac{DC}{b}, \qquad\text{so}\qquad BD : DC = c : b .

Figure 5.36. The perpendicular from

C

divides

AB

into

b\cos A

and

a\cos B

; the bisector

AD

divides

BC

in the ratio

c : b

Points Associated with a Triangle

The following geometric properties of a triangle should also be familiar.

The perpendicular bisectors of the sides of a triangle $ABC$ meet at a point called the circumcentre, which is the centre of the circle through $A$ , $B$ and $C$ , the circumcircle. Its radius is the $R$ of the sine rule.
The bisectors of the angles $A$ , $B$ and $C$ meet at a point called the incentre, which is the centre of the circle that touches all three sides, the inscribed circle.
The altitudes meet at a point $H$ called the orthocentre.
The medians meet at a point $G$ called the centroid.

Figure 5.37. The circumcentre, incentre, orthocentre and centroid of a triangle.

The Area of a Triangle

The area of a triangle can be found using any of the following.

Half the base times the perpendicular height.
$\frac12 ab\sin C$ , or the corresponding $\frac12 bc\sin A$ and $\frac12 ca\sin B$ . This follows from the first, since the height of $A$ above the side $CB$ , of length $a$ , is $b\sin C$ .
$\sqrt{s(s - a)(s - b)(s - c)}$ , where $s = \frac12(a + b + c)$ , which we accept here.

Example 5.63.

In a surveying exercise, $P$ and $Q$ are two points on land which is inaccessible. To find the distance $PQ$ , a line $AB$ of length $200$ metres is drawn so that $P$ and $Q$ are on opposite sides of $AB$ . The following angles are measured:

\angle ABP = 60^\circ, \qquad \angle ABQ = 46^\circ, \qquad \angle BAP = 30^\circ, \qquad \angle BAQ = 67^\circ .

Find the distance $PQ$ .

In the triangle $APB$ , $\angle APB = 90^\circ$ , hence $AP = 200\cos 30^\circ = 173.2$ m. In the triangle $ABQ$ , $\angle AQB = 67^\circ$ , hence

\frac{AQ}{\sin 46^\circ} = \frac{200}{\sin 67^\circ}, \qquad AQ = 156.3 \text{ m}.

Then in the triangle $APQ$ , where $\angle PAQ = 30^\circ + 67^\circ = 97^\circ$ , the cosine rule gives

PQ^2 = AP^2 + AQ^2 - 2 \cdot AP \cdot AQ\cos 97^\circ = 173.2^2 + 156.3^2 - 2(173.2)(156.3)(-0.1219),

hence $PQ = 247$ m.

Figure 5.38. The baseline

AB

and the inaccessible points

P

and

Q

Example 5.64.

In a triangle $ABC$ , $BC = 7.4$ cm, $AC = 4.1$ cm and $\angle ACB = 66^\circ$ . Calculate the other angles of the triangle, the area of the triangle, and the distance of the incentre $I$ of the triangle from $BC$ .

Using the cosine rule, $c^2 = 7.4^2 + 4.1^2 - 2(7.4)(4.1)\cos 66^\circ$ gives $c = 6.85$ cm. Then the sine rule, $\dfrac{7.4}{\sin A} = \dfrac{4.1}{\sin B} = \dfrac{6.85}{\sin 66^\circ}$ , gives $A = 80.8^\circ$ and $B = 33.2^\circ$ ; $A$ is acute since $a^2 < b^2 + c^2$ .
The area of the triangle is $\frac12 ab\sin C = \frac12(7.4)(4.1)\sin 66^\circ = 13.86\ \text{cm}^2$ .
The incentre $I$ is the centre of the inscribed circle, and is therefore at the same distance $r$ from all three sides. Joining $AI$ , $BI$ and $CI$ divides the triangle into three triangles, $AIB$ , $BIC$ and $CIA$ , with areas $\frac12 cr$ , $\frac12 ar$ and $\frac12 br$ . So the total area of the triangle $ABC$ is $\frac12 r(a + b + c) = 9.175r$ . But this area is $13.86\ \text{cm}^2$ , hence the distance of $I$ from $BC$ is $r = 1.5$ cm.

Figure 5.39. The incentre

I

at the distance

r

from each side, dividing the triangle into three.

Problem 5.20.

Calculate the area of the triangle whose sides are $11$ , $10$ and $15$ .
In a triangle $PQR$ , $p = 2.8$ m and $q = 4.5$ m. If the area of the triangle is $5.84\ \text{m}^2$ , find the two possible values of $r$ .
Show that, for any triangle $ABC$ , $abc = 4\Delta R$ , where $\Delta$ is the area of the triangle and $R$ is the radius of its circumcircle.
$ABC$ is a triangle and $D$ is the point on $BC$ for which $BD = DC$ . The angle $BAD$ is $20^\circ$ and the angle $CAD$ is $30^\circ$ . Find the angle $ACB$ .

Exercises

Questions marked with an examining board are taken from past A-level papers: JMB is the Joint Matriculation Board, U of L the University of London, C Cambridge, and AEB the Associated Examining Board.

Exercise 5.1.

A chord of a circle subtends an angle of $\theta$ radians at the centre of the circle. If the area of the minor segment cut off by the chord is one sixth of the area of the circle, show that

\sin\theta = \theta - \frac{\pi}{3} .

Exercise 5.2.

$P$ and $Q$ are points on a circle of radius $r$ , and the chord $PQ$ subtends an angle of $2\theta$ radians at the centre $O$ . If $A$ is the area enclosed by the minor arc $PQ$ and the chord $PQ$ , and $B$ is the area enclosed by the arc $PQ$ and the tangents to the circle at $P$ and $Q$ , show that

A - B = r^2(2\theta - \tan\theta - \sin\theta\cos\theta) .

Exercise 5.3.

Three cylinders are placed in contact with each other with their axes parallel. The radii of the cylinders are $3$ cm, $4$ cm and $5$ cm. An elastic band is stretched round the three cylinders so that the plane of the band is perpendicular to the axes of the cylinders. Calculate the length of the part of the band in contact with the largest cylinder. (U of L)

Exercise 5.4.

Without the use of a calculator, find, for each of the following equations, all the solutions in the interval $0^\circ \leqslant x \leqslant 180^\circ$ .

$\cos(x + 30^\circ) = \cos(60^\circ - 3x)$ ;
$\sin(x + 20^\circ) = \cos 3x$ . (JMB)

Exercise 5.5.

Find the general solution of the equation $\cos 3x + \cos x = \sin 2x$ . (U of L)

Exercise 5.6.

If $\sin(\theta - \alpha) = k\sin(\theta + \alpha)$ , find $\tan\theta$ in terms of $\tan\alpha$ and $k$ , and so determine the possible values of $\theta$ between $0^\circ$ and $360^\circ$ when $k = \frac12$ and $\alpha = 150^\circ$ .
Show, without the use of a calculator, that $x = \dfrac{\pi}{10}$ satisfies the equation $\cos 3x = \sin 2x$ . By expressing this equation in terms of $\sin x$ and $\cos x$ , show that $\sin\dfrac{\pi}{10}$ is a root of the equation $4s^2 + 2s - 1 = 0$ . (C)

Exercise 5.7.

Find all the values of $\theta$ in the range $0 \leqslant \theta \leqslant 2\pi$ for which $\sin\theta + \sin 3\theta = \cos\theta + \cos 3\theta$ . (JMB)

Exercise 5.8.

Find, to the nearest minute, the acute angle $\alpha$ for which $4\cos\theta - 3\sin\theta = 5\cos(\theta + \alpha)$ . Calculate the values of $\theta$ in the interval $-180^\circ < \theta < 180^\circ$ for which the function $f(\theta) = 4\cos\theta - 3\sin\theta - 4$ attains its greatest value, its least value and the value zero. (JMB)

Exercise 5.9.

Show that $(\sin 2\theta - \sin\theta)(1 + 2\cos\theta) = \sin 3\theta$ .
Find the values of $x$ between $0^\circ$ and $360^\circ$ which satisfy the equation $\sin x + \sin 2x = \sin 3x$ . (C)

Exercise 5.10.

Show that $\sec x + \tan x = \tan\left(\dfrac{\pi}{4} + \dfrac{x}{2}\right)$ , and deduce a similar expression for $\sec x - \tan x$ . Hence find in surd form the values of $\tan\dfrac{\pi}{12}$ and $\tan\dfrac{5\pi}{12}$ . (AEB, 1975)

Exercise 5.11.

Show that $\cos 3\theta - \sin 3\theta = (\cos\theta + \sin\theta)(1 - 4\cos\theta\sin\theta)$ .
Show that if $\sec A = \cos B + \sin B$ , then $\tan^2 A = \sin 2B$ and $\cos 2A = \tan^2\left(\dfrac{\pi}{4} - B\right)$ . (C)

Exercise 5.12.

By expressing $\sec 2x$ and $\tan 2x$ in terms of $\tan x$ , or otherwise, solve the equation $2\tan x + \sec 2x = 2\tan 2x$ , giving all the solutions between $-180^\circ$ and $180^\circ$ . (U of L)

Exercise 5.13.

Express $\sqrt3\sin\theta - \cos\theta$ in the form $R\sin(\theta - \alpha)$ , where $R$ is positive. Find all the values of $\theta$ in the range $0^\circ \leqslant \theta \leqslant 360^\circ$ which satisfy the equation $4\sin\theta\cos\theta = \sqrt3\sin\theta - \cos\theta$ . (JMB)

Exercise 5.14.

Show that $(\cot\theta + \operatorname{cosec}\theta)^2 = \dfrac{1 + \cos\theta}{1 - \cos\theta}$ , and hence, or otherwise, solve the equation $(\cot 2\theta + \operatorname{cosec} 2\theta)^2 = \sec 2\theta$ for values of $\theta$ between $0^\circ$ and $180^\circ$ .
Find the general solution of the equation $\sin 2x + \sin 3x + \sin 5x = 0$ . (AEB, 1973)

Exercise 5.15.

For $\theta\not\equiv\pi\pmod{2\pi}$ , use the formulae expressing $\sin\theta$ and $\cos\theta$ in terms of $t = \tan\dfrac{\theta}{2}$ , or otherwise, to show that

\frac{1 + \sin\theta}{5 + 4\cos\theta} = \frac{(1 + t)^2}{9 + t^2} .

Check the omitted values of $\theta$ directly, and deduce that $0 \leqslant \dfrac{1 + \sin\theta}{5 + 4\cos\theta} \leqslant \dfrac{10}{9}$ for all values of $\theta$ . (C)

Exercise 5.16.

Show that $\cos^4 x + \sin^4 x = 1 - \frac12\sin^2 2x$ .
Solve, for $0^\circ \leqslant x \leqslant 180^\circ$ , the equation $\sin x + \sin 5x = \sin 3x$ .
Find the general solution of the equation $3\cos x + 4\sin x = 2$ . (U of L)

Exercise 5.17.

Show that $\operatorname{cosec}\theta + \cot\theta = \cot\dfrac{\theta}{2}$ . Hence

deduce the values, in surd form, of $\cot\dfrac{\pi}{8}$ and $\cot\dfrac{\pi}{12}$ ;
express $\operatorname{cosec}\theta + \operatorname{cosec} 2\theta + \operatorname{cosec} 4\theta$ as the difference of two cotangents;
show, without using a calculator, that $\operatorname{cosec}\dfrac{4\pi}{15} + \operatorname{cosec}\dfrac{8\pi}{15} + \operatorname{cosec}\dfrac{16\pi}{15} + \operatorname{cosec}\dfrac{32\pi}{15} = 0$ . (C)

Exercise 5.18.

Express $7\sin x - 24\cos x$ in the form $R\sin(x - \alpha)$ , where $R$ is positive and $\alpha$ is an acute angle. Hence, or otherwise, solve the equation $7\sin x - 24\cos x = 15$ for $0^\circ \leqslant x \leqslant 360^\circ$ .
Solve the simultaneous equations $\cos x + \cos y = 1$ and $\sec x + \sec y = 4$ for $0^\circ \leqslant x \leqslant 180^\circ$ and $0^\circ \leqslant y \leqslant 180^\circ$ . (AEB, 1976)

Exercise 5.19.

Express $\cos 2x - \sin 2x$ in the form $R\cos(2x + \alpha)$ , giving the values of $R$ and $\alpha$ . Hence find the general solution of each of the following equations.

$\cos 2x - \sin 2x = 1$ ;
$\cos 2x - \sin 2x = \sqrt2\cos 4x$ . (U of L)

Exercise 5.20.

Find, in the range $-180^\circ \leqslant x \leqslant 180^\circ$ , the solutions of the equation $\cos 5x = \cos x$ .
Show that $\dfrac{1 + \cos\theta + \sin\theta}{1 - \cos\theta + \sin\theta} = \dfrac{1 + \cos\theta}{\sin\theta}$ . (JMB)

Exercise 5.21.

If $\sin\theta + \sin 2\theta + \sin 3\theta + \sin 4\theta = 0$ , show that $\theta$ is either a multiple of $\dfrac{\pi}{2}$ or a multiple of $\dfrac{2\pi}{5}$ . (U of L)

Exercise 5.22.

Write down the expansions of $\cos(A + B)$ and $\cos(A - B)$ in terms of the cosines and sines of $A$ and $B$ .

Find angles $x$ and $y$ , each between $0$ and $90^\circ$ , which satisfy the simultaneous equations $\cos x\cos y = 0.6$ and $\sin x\sin y = 0.2$ .
Show that $\cos 3x = 4\cos^3 x - 3\cos x$ . Hence find all the solutions, in the range $-180^\circ \leqslant x \leqslant 180^\circ$ , of the equation $2\cos 3x + \cos 2x + 1 = 0$ . (JMB)

Exercise 5.23.

By means of the substitution $\tan\theta = t$ , or otherwise, find the values of $\theta$ in the range $0 < \theta < \dfrac{\pi}{2}$ such that

(2 - \tan\theta)(1 + \sin 2\theta) - 2 = 0 .

Show that, when $\theta$ is small, $(2 - \tan\theta)(1 + \sin 2\theta) - 2 \approx 3\theta$ . (JMB)

Exercise 5.24.

A quadrilateral $ABCD$ is right-angled at $B$ and at $D$ , and the angle $DAB$ is $132^\circ$ . If $DA = 4$ and $AB = 7$ , find the lengths of the diagonals of the quadrilateral and the radius of the inscribed circle of the triangle $ABC$ . (U of L)

Exercise 5.25.

Three towns $A$ , $B$ and $C$ are all at sea level. The bearings of the towns $B$ and $C$ from $A$ , measured clockwise from north, are $36^\circ$ and $247^\circ$ respectively. If $B$ is $120$ km from $A$ and $C$ is $234$ km from $A$ , calculate the distance and the bearing of the town $B$ from $C$ . (AEB, 1972)

Check Yourself

Fresh questions on the whole chapter — none of them is worked out above. Do each on paper first; the box only tells you whether you got there.

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Measuring Angles

Measurement of Rotation

The Degree

The Radian

The Circular Functions

The Ratios of an Acute Angle

Length of an Arc and Area of a Sector

The Ratios of a General Angle

Negative Angles

Solving Triangles

The Sine Rule

The Ambiguous Case

The Cosine Rule

The Graphs of the Circular Functions

The Sine Function

The Cosine Function

The Tangent Function

Special Values

The Reciprocal Ratios

Inverse Circular Functions

Common Trigonometric Ratios

Complementary Angles

Trigonometric Equations

Principal Values

Secondary Values

Solutions in a Specified Range

General Solutions

Multiple Angles

The Equation cos⁡A=cos⁡B\cos A = \cos BcosA=cosB

Graphs of Multiple and Compound Angles

Trigonometric Identities

The Pythagorean Identities

Compound Angle Identities

Double Angle Identities

Identities and the Inverse Functions

The Half Angle Identities

The Expression acos⁡θ+bsin⁡θa\cos\theta + b\sin\thetaacosθ+bsinθ

The Graph of acos⁡θ+bsin⁡θa\cos\theta + b\sin\thetaacosθ+bsinθ

The Equation acos⁡θ+bsin⁡θ=ca\cos\theta + b\sin\theta = cacosθ+bsinθ=c

The Factor Formulae

Small Angles

Further Properties of Triangles

The Cotangent Formula

Other Relationships

The Projection Formula

The Difference of Two Sides

The Bisector of an Angle

Points Associated with a Triangle

The Area of a Triangle

Exercises

Check Yourself

The Equation $\cos A = \cos B$

The Expression $a\cos\theta + b\sin\theta$

The Graph of $a\cos\theta + b\sin\theta$

The Equation $a\cos\theta + b\sin\theta = c$