1.05PureCore

Trigonometric Ratios

Q: Why use the unit circle?

It defines sine, cosine and tangent for every angle, not just acute ones, and makes the signs and symmetries visible.

Q: When does the ambiguous case happen?

With two sides and a non-included angle (SSA), a sine-rule calculation can give two valid triangles. Find the principal angle, test its supplement, and retain each only if all angles are positive and their sum can be 180 degrees. Two solutions are not guaranteed.

Q: Why do radians matter?

They make arc length, sector area and the derivatives of trigonometric functions simple. Calculus with trig functions only works cleanly in radians.

Q: Sine rule or cosine rule?

Cosine rule for two sides and the included angle, or three sides. Sine rule when you have a matching side and opposite angle.

At GCSE, sine, cosine and tangent were ratios of sides in a right-angled triangle, which only makes sense for angles between $0^{\circ}$ and $90^{\circ}$ . A-Level redefines them using a point moving round a circle, so that $\sin 210^{\circ}$ or $\cos(-\tfrac{\pi}{3})$ mean something — and in doing so it explains the shapes of the graphs, the symmetries between angles, and the rules for non-right-angled triangles.

45 min Trigonometry

The big picture

The unit circle is the idea that ties this whole chapter together. Put a point on a circle of radius 1 at angle $\theta$ from the positive $x$ -axis; its coordinates are $(\cos\theta, \sin\theta)$ and the gradient of the radius is $\tan\theta$ . Every sign rule, every symmetry like $\sin(180^{\circ} - \theta) = \sin\theta$ , and every graph follows from where that point is. The sine and cosine rules then extend right-angled trigonometry to any triangle. Radians, the natural angle measure, replace degrees in almost all later work because they make the calculus of trigonometric functions clean.

What you'll be able to do

Define sine, cosine and tangent for any angle using the unit circle
Recall exact values and use symmetry to find ratios of related angles
Sketch the graphs of $\sin\theta$ , $\cos\theta$ and $\tan\theta$
Use the sine rule, cosine rule and $\tfrac12 ab\sin C$ , including the ambiguous case
Convert between degrees and radians and use arc length and sector area

The unit circle definition

Take the point $P$ on the circle $x^{2} + y^{2} = 1$ at angle $\theta$ measured anticlockwise from the positive $x$ -axis. Define $\cos\theta$ as its $x$ -coordinate, $\sin\theta$ as its $y$ -coordinate, and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ .

For $0 < \theta < 90^{\circ}$ this agrees with SOH CAH TOA, because the hypotenuse is 1. But now any angle works, including negative angles (measured clockwise) and angles beyond $360^{\circ}$ .

The signs follow from the quadrant: all three ratios are positive in the first quadrant, only sine in the second, only tangent in the third, and only cosine in the fourth — often remembered as CAST.

Why this works

Because $P$ lies on $x^{2} + y^{2} = 1$ , it is immediate that $\cos^{2}\theta + \sin^{2}\theta = 1$ for every angle. The most important identity in the course is just Pythagoras on the unit circle.

Exact values and symmetry

Learn the exact values for $30^{\circ}$ , $45^{\circ}$ and $60^{\circ}$ : $\sin 30^{\circ} = \tfrac12$ , $\sin 45^{\circ} = \tfrac{\sqrt2}{2}$ , $\sin 60^{\circ} = \tfrac{\sqrt3}{2}$ , with cosines in the reverse order, and $\tan 30^{\circ} = \tfrac{\sqrt3}{3}$ , $\tan 45^{\circ} = 1$ , $\tan 60^{\circ} = \sqrt3$ .

Symmetry of the unit circle gives related angles: $\sin(180^{\circ} - \theta) = \sin\theta$ , $\cos(180^{\circ} - \theta) = -\cos\theta$ , $\cos(-\theta) = \cos\theta$ , $\sin(-\theta) = -\sin\theta$ , and $\tan(180^{\circ} + \theta) = \tan\theta$ .

Example 1Find the exact value of

\cos 300^{\circ}

and of

\tan 135^{\circ}

300^{\circ}

is in the fourth quadrant,

60^{\circ}

below the axis; cosine is positive there.

2So

\cos 300^{\circ} = \cos 60^{\circ} = \tfrac12

135^{\circ}

is in the second quadrant,

45^{\circ}

from the axis; tangent is negative there.

Answer

\cos 300^{\circ} = \tfrac12

\tan 135^{\circ} = -1

Graphs of the ratios

$y = \sin\theta$ and $y = \cos\theta$ are waves with $\textbf{period}$ $360^{\circ}$ ( $2\pi$ ) and range $-1 \le y \le 1$ . The cosine graph is the sine graph shifted $90^{\circ}$ left.

$y = \tan\theta$ has period $180^{\circ}$ ( $\pi$ ), takes every real value, and has vertical asymptotes where $\cos\theta = 0$ : at $\theta = 90^{\circ}, 270^{\circ}, \dots$

Transformations follow the usual rules: $y = 3\sin 2\theta$ has amplitude 3 and period $180^{\circ}$ ; $y = \cos(\theta - 30^{\circ})$ is $y = \cos\theta$ translated $30^{\circ}$ right.

Tip — For $y = \sin k\theta$ the period is $\dfrac{360^{\circ}}{k}$ . A larger $k$ squashes the graph horizontally.

Non-right-angled triangles

Label sides $a$ , $b$ , $c$ opposite angles $A$ , $B$ , $C$ . The $\textbf{sine rule}$ links pairs of sides and opposite angles; the $\textbf{cosine rule}$ links all three sides with one angle.

Use the cosine rule when you know two sides and the included angle, or all three sides. Use the sine rule when you know a side and its opposite angle.

$\textbf{The ambiguous case}$ : finding an angle with the sine rule can give two answers, because $\sin\theta = \sin(180^{\circ} - \theta)$ . Both are valid if the second angle still leaves room for a triangle.

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

Sine rule.

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

Cosine rule.

\text{Area} = \tfrac12 ab\sin C

Area of any triangle.

Example 1In triangle

ABC

a = 7

cm,

b = 9

cm and

A = 40^{\circ}

. Find the possible values of

B

\sin B = \dfrac{9\sin 40^{\circ}}{7} = 0.8264

B = 55.7^{\circ}

B = 180^{\circ} - 55.7^{\circ} = 124.3^{\circ}

3Check the second:

40^{\circ} + 124.3^{\circ} = 164.3^{\circ} < 180^{\circ}

, so it is possible.

Answer

B = 55.7^{\circ}

124.3^{\circ}

Radians

One $\textbf{radian}$ is the angle at the centre of a circle subtended by an arc equal in length to the radius. A full turn is $2\pi$ radians, so $180^{\circ} = \pi$ radians.

In radians, the arc length and sector area formulas are especially simple, as shown below.

Common conversions: $30^{\circ} = \tfrac{\pi}{6}$ , $45^{\circ} = \tfrac{\pi}{4}$ , $60^{\circ} = \tfrac{\pi}{3}$ , $90^{\circ} = \tfrac{\pi}{2}$ .

s = r\theta, \qquad A = \tfrac12 r^{2}\theta

Arc length and sector area,

\theta

in radians.

Example 1A sector has radius 8 cm and angle

\tfrac{3\pi}{4}

. Find its perimeter and area in terms of

\pi

1Arc:

8 \times \tfrac{3\pi}{4} = 6\pi

2Perimeter:

6\pi + 2 \times 8

3Area:

\tfrac12 \times 64 \times \tfrac{3\pi}{4} = 24\pi

AnswerPerimeter

6\pi + 16

cm; area

24\pi

cm²

Tip — Check your calculator mode before every trigonometry question. A radian answer computed in degree mode is the most avoidable lost mark in the paper.

Think like an examiner

Use the unit circle or CAST to justify signs, not memory alone.
Give exact values in surd form when the question says "exact".
Consider the ambiguous case whenever the sine rule is used to find an angle.
Use radians for arc length and sector area.

Common misconceptions

Giving

\cos 120^{\circ} = \tfrac12

Cosine is negative in the second quadrant:

-\tfrac12

Taking only the calculator value of

B

from the sine rule.

Check whether

180^{\circ} - B

also gives a valid triangle.

Using

s = r\theta

with

\theta

in degrees.

Convert to radians first.

Stating the period of

\tan\theta

360^{\circ}

Tangent repeats every

180^{\circ}

Trigonometric ratios

\tan\theta = \dfrac{\sin\theta}{\cos\theta}

Tangent.

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

Cosine rule.

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

Sine rule.

\pi \text{ rad} = 180^{\circ}

Radians.

Stretch yourself

Two ships leave port $P$ at the same time. Ship $A$ sails 12 km on a bearing of $050^{\circ}$ and ship $B$ sails 17 km on a bearing of $160^{\circ}$ . Find the distance between the ships and the bearing of $B$ from $A$ .

Hint — The angle $APB$ is the difference of the bearings. Use the cosine rule for $AB$ , then the sine rule for angle $PAB$ , and turn that into a bearing at $A$ .

Reveal solution

SolutionAngle

APB = 160^{\circ} - 50^{\circ} = 110^{\circ}

. Cosine rule:

AB^{2} = 12^{2} + 17^{2} - 2(12)(17)\cos 110^{\circ} = 144 + 289 + 139.54 = 572.54

, so

AB = 23.93

km. Sine rule:

\sin \angle PAB = \dfrac{17\sin 110^{\circ}}{23.93} = 0.6676

, so

\angle PAB = 41.9^{\circ}

(the obtuse option is impossible, because the triangle already contains

110^{\circ}

). At

A

, the bearing back to

P

050^{\circ} + 180^{\circ} = 230^{\circ}

. Turning from

AP

towards

B

is anticlockwise when viewed from above, since

B

lies to the east of the line

AP

, so the bearing of

B

from

A

230^{\circ} - 41.9^{\circ} = 188.1^{\circ}

. So

AB \approx 23.9

km on a bearing of about

188^{\circ}

\blacksquare

Questions students ask

Why use the unit circle?

It defines sine, cosine and tangent for every angle, not just acute ones, and makes the signs and symmetries visible.

When does the ambiguous case happen?

With two sides and a non-included angle (SSA), a sine-rule calculation can give two valid triangles. Find the principal angle, test its supplement, and retain each only if all angles are positive and their sum can be 180 degrees. Two solutions are not guaranteed.

Why do radians matter?

They make arc length, sector area and the derivatives of trigonometric functions simple. Calculus with trig functions only works cleanly in radians.

Sine rule or cosine rule?

Cosine rule for two sides and the included angle, or three sides. Sine rule when you have a matching side and opposite angle.

Key takeaways

On the unit circle, $P = (\cos\theta, \sin\theta)$ and $\tan\theta = \sin\theta/\cos\theta$ .
Signs follow CAST; related angles follow from symmetry.
Sine and cosine have period $360^{\circ}$ ; tangent has period $180^{\circ}$ with asymptotes.
Sine rule, cosine rule and $\tfrac12 ab\sin C$ solve any triangle; watch the ambiguous case.
$\pi$ radians $= 180^{\circ}$ ; $s = r\theta$ and $A = \tfrac12 r^{2}\theta$ .