Sine and cosine beyond 90° · Trigonometric ratios

A right-angled triangle can’t have an angle of $150^\circ$ , yet you need $\sin 150^\circ$ for the sine rule, for bearings and for trig graphs. The idea that extends sine and cosine to any angle is a circle.

The unit circle

Take a circle of radius 1 with its centre at the origin. Start from the positive $x$ -axis and turn anticlockwise through an angle $\theta$ . The point you reach has:

an across distance ( $x$ -coordinate) of $\cos\theta$ ;
an up distance ( $y$ -coordinate) of $\sin\theta$ .

For acute angles this is exactly the right-angled triangle from lesson 1, with hypotenuse 1. For bigger angles it keeps going. Because the point stays on a circle of radius 1, $\sin\theta$ and $\cos\theta$ always lie between $-1$ and $1$ : that is what questions about the greatest or least value of an expression in $\cos\theta$ use.

Sine and cosine for any angleDrag the point round the circle

150°angle θ−0.866cos θ (across): negative0.5sin θ (up): positive−0.577tan θ = sin ÷ cos

In this quadrant only sine is positive. The acute angle to the x-axis, the reference angle, is 30°, so sin 150° = sin 30°, cos 150° = −cos 30°. Work out the size from the reference angle, then give it the sign of the quadrant.

Go round slowly and watch the signs. Across is negative on the left half, and up is negative on the bottom half.

Key idea

Quadrant	Angles	Positive
First	$0^\circ$ to $90^\circ$	All
Second	$90^\circ$ to $180^\circ$	Sin only
Third	$180^\circ$ to $270^\circ$	Tan only
Fourth	$270^\circ$ to $360^\circ$	Cos only

Going anticlockwise from the first quadrant, the letters spell A S T C.

Which ratio is positiveA S T C, anticlockwise from 0°

The reference angle

Every angle has an acute reference angle: its angle to the $x$ -axis. The sine and cosine have the same size as those of the reference angle; only the sign changes, according to the quadrant.

$\theta$ in the…	Reference angle
second quadrant	$180^\circ - \theta$
third quadrant	$\theta - 180^\circ$
fourth quadrant	$360^\circ - \theta$

So $\sin 150^\circ = \sin 30^\circ = \frac12$ (sine is positive in the second quadrant), and $\cos 135^\circ = -\cos 45^\circ = -\frac{\sqrt2}{2}$ (cosine is negative there).

A past question, step by step

Worked example · WAEC 2021

WAEC 2021 · Paper 1 · Q22

If $\tan\theta = \frac34$ and $180^\circ < \theta < 270^\circ$ , find the value of $\cos\theta$ .

Find the size, ignoring signs
Sketch a right-angled triangle with opposite 3 and adjacent 4. The hypotenuse is $\sqrt{3^2 + 4^2} = 5$ . So the size of $\cos\theta$ is $\frac45$ .
Think first. $\tan\theta = \frac34$ gives the opposite and adjacent. What is the hypotenuse?
Find the sign from the quadrant
$180^\circ < \theta < 270^\circ$ is the third quadrant, where only tangent is positive.
Think first. $\theta$ is between $180^\circ$ and $270^\circ$ . Which quadrant is that, and is cosine positive there?
Put them together
Cosine is negative in the third quadrant, so $\cos\theta = -\frac45$ . The answer is C. (Check: $\tan\theta = \frac34$ is positive, as it should be in the third quadrant.)

Your turn

WAEC 2020 · Paper 1 · Q27

Trigonometric ratios

If $\tan y$ is positive and $\sin y$ is negative, in which quadrant would $y$ lie?

Worked solution (try it first)

Sine is negative in the third and fourth quadrants.
Tangent is positive in the first and third quadrants.
Only the third quadrant is on both lists, so the answer is the third only, option C.

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