Mathematics 9709 · Pure Mathematics 1

Exact values and angles of any size

The exam uses angles bigger than $90^\circ$ . It uses $120^\circ$ , $240^\circ$ , and even angles below zero.
So you need two things. You need the exact values of the special angles. You also need a way to get $\sin,\cos,\tan$ of any angle. One picture gives you both. It is called the unit circle.

What sin, cos and tan really mean

Forget triangles: take a circle of radius 1 and turn a point on it by an angle $\theta$ .

sin θ = how high the point is (height above the middle).
cos θ = how far across it is (left or right of the middle).
tan θ = the slope of the line out to the point: $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\text{height}}{\text{across}}$ .

Drag the angle below. Watch the height and the across change. The height draws the wave on the right: the sine graph is just the height drawn against the angle.

angle θ

θ = 50°·sin = height = 0.77·cos = across = 0.64·tan = slope = 1.19

A sine or cosine is never more than 1 and never less than −1. The point never leaves the circle. This one fact stops many wrong answers.

For any angle, each ratio (sine, cosine or tangent) is the exact value of a small acute angle, plus a plus or minus sign from the quadrant. Get these two pieces. Then you never need a calculator for a special angle.

The special angles you must learn by heart

Cut an equilateral triangle (side 2) in half. This gives $30^\circ$ and $60^\circ$ .
A right-angled triangle with two equal short sides gives $45^\circ$ .

These two triangles give every exact value you need:

$\sin30^\circ=\tfrac12,\ \cos30^\circ=\tfrac{\sqrt3}{2},\ \tan30^\circ=\tfrac{1}{\sqrt3}$

$\sin45^\circ=\cos45^\circ=\tfrac{1}{\sqrt2},\ \tan45^\circ=1$

$\sin60^\circ=\tfrac{\sqrt3}{2},\ \cos60^\circ=\tfrac12,\ \tan60^\circ=\sqrt3$

Easy way to remember: $\sin$ goes up the list $\tfrac12,\ \tfrac{1}{\sqrt2},\ \tfrac{\sqrt3}{2}$ as the angle gets bigger.
$\cos$ uses the same three numbers in the other order.

Angles of any size: the unit circle and CAST

Turn a point $P$ on the unit circle by $\theta$ . Then $P=(\cos\theta,\ \sin\theta)$ . So cos is the across, and sin is the height.
This gives the ratios for every angle, plus or minus. The plus or minus signs come from the CAST diagram.

The CAST diagram: which ratio is positive in each quadrant. Go anticlockwise from 0°.

Read “All, Sin, Tan, Cos” round from the bottom-right.
A ( $0$ – $90^\circ$ ): all are positive. Then only Sine. Then only Tan. Then only Cos.

Find the basic (acute) angle to the $x$ -axis. Take the exact value of that angle. Then add the plus or minus sign that CAST gives you. That is the whole method: basic angle plus sign.

Worked examples

Worked example

Find the exact value of $\sin 120^\circ$

$120^\circ$ lands in the second quadrant. Its basic angle to the horizontal is $180^\circ-120^\circ=60^\circ$ .
CAST: in the second quadrant S says sine is positive, so $\sin120^\circ=+\sin60^\circ$ .
Read off the exact value of $\sin60^\circ$ .

Answer

\sin120^\circ=\tfrac{\sqrt3}{2}

Worked example

Three AP terms are $2\cos x$ , $-6\sqrt3\sin x$ , $10\cos x$ : find exact $x$ $(\tfrac12\pi < x < \pi)$

This is 9709/12 Nov 2021 Q5. In an arithmetic progression the middle term is the average of the two terms next to it, so $2\times(-6\sqrt3\sin x)=2\cos x+10\cos x$ :

Simplify both sides: $-12\sqrt3\,\sin x=12\cos x$ .
Divide by $\cos x$ to make a single tangent (here it's safe — the working forces it): $\tan x=-\tfrac{1}{\sqrt3}$ .
Basic angle with $\tan=\tfrac{1}{\sqrt3}$ is $30^\circ=\tfrac{\pi}{6}$ . We need the second quadrant (where $\tan$ is negative and $\tfrac{\pi}{2}<x<\pi$ ): take $\pi-\tfrac{\pi}{6}$ .

Answer

x=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}

Where the marks go

Common mistake

When the question says exact value or surd form, never round.

\tfrac{\sqrt3}{2}

is the answer;

0.866

earns nothing.

Common mistake

The basic angle is always measured to the

x

-axis, not the

y

-axis. For

120^\circ

that is

180-120=60^\circ

, not

30^\circ

Now you try

Without a calculator, find the exact value of $\cos\tfrac{7\pi}{6}$ . (9709-style)

Your turn— tap to reveal the worked answer (third quadrant, basic angle π/6)

$\tfrac{7\pi}{6}$ is in the third quadrant. The basic angle is $\tfrac{7\pi}{6}-\pi=\tfrac{\pi}{6}$ . There only Tan is positive, so the cosine is negative:

Answer

\cos\tfrac{7\pi}{6}=-\cos\tfrac{\pi}{6}=-\tfrac{\sqrt3}{2}

PreviousApplications: shaded-region problems NextGraphs of sin, cos and tan

Exact values and angles of any size

The exam uses angles bigger than $90^\circ$ . It uses $120^\circ$ , $240^\circ$ , and even angles below zero.
So you need two things. You need the exact values of the special angles. You also need a way to get $\sin,\cos,\tan$ of any angle. One picture gives you both. It is called the unit circle.

What sin, cos and tan really mean

Forget triangles: take a circle of radius 1 and turn a point on it by an angle $\theta$ .

sin θ = how high the point is (height above the middle).
cos θ = how far across it is (left or right of the middle).
tan θ = the slope of the line out to the point: $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\text{height}}{\text{across}}$ .

Drag the angle below. Watch the height and the across change. The height draws the wave on the right: the sine graph is just the height drawn against the angle.

angle θ

θ = 50°·sin = height = 0.77·cos = across = 0.64·tan = slope = 1.19

A sine or cosine is never more than 1 and never less than −1. The point never leaves the circle. This one fact stops many wrong answers.

The special angles you must learn by heart

Cut an equilateral triangle (side 2) in half. This gives $30^\circ$ and $60^\circ$ .
A right-angled triangle with two equal short sides gives $45^\circ$ .

These two triangles give every exact value you need:

$\sin30^\circ=\tfrac12,\ \cos30^\circ=\tfrac{\sqrt3}{2},\ \tan30^\circ=\tfrac{1}{\sqrt3}$

$\sin45^\circ=\cos45^\circ=\tfrac{1}{\sqrt2},\ \tan45^\circ=1$

$\sin60^\circ=\tfrac{\sqrt3}{2},\ \cos60^\circ=\tfrac12,\ \tan60^\circ=\sqrt3$

Easy way to remember: $\sin$ goes up the list $\tfrac12,\ \tfrac{1}{\sqrt2},\ \tfrac{\sqrt3}{2}$ as the angle gets bigger.
$\cos$ uses the same three numbers in the other order.

Angles of any size: the unit circle and CAST

Turn a point $P$ on the unit circle by $\theta$ . Then $P=(\cos\theta,\ \sin\theta)$ . So cos is the across, and sin is the height.
This gives the ratios for every angle, plus or minus. The plus or minus signs come from the CAST diagram.

The CAST diagram: which ratio is positive in each quadrant. Go anticlockwise from 0°.

Read “All, Sin, Tan, Cos” round from the bottom-right.
A ( $0$ – $90^\circ$ ): all are positive. Then only Sine. Then only Tan. Then only Cos.

Find the basic (acute) angle to the $x$ -axis. Take the exact value of that angle. Then add the plus or minus sign that CAST gives you. That is the whole method: basic angle plus sign.

Worked examples

Worked example

Find the exact value of $\sin 120^\circ$

$120^\circ$ lands in the second quadrant. Its basic angle to the horizontal is $180^\circ-120^\circ=60^\circ$ .
CAST: in the second quadrant S says sine is positive, so $\sin120^\circ=+\sin60^\circ$ .
Read off the exact value of $\sin60^\circ$ .

Answer

\sin120^\circ=\tfrac{\sqrt3}{2}

Worked example

Three AP terms are $2\cos x$ , $-6\sqrt3\sin x$ , $10\cos x$ : find exact $x$ $(\tfrac12\pi < x < \pi)$

This is 9709/12 Nov 2021 Q5. In an arithmetic progression the middle term is the average of the two terms next to it, so $2\times(-6\sqrt3\sin x)=2\cos x+10\cos x$ :

Simplify both sides: $-12\sqrt3\,\sin x=12\cos x$ .
Divide by $\cos x$ to make a single tangent (here it's safe — the working forces it): $\tan x=-\tfrac{1}{\sqrt3}$ .
Basic angle with $\tan=\tfrac{1}{\sqrt3}$ is $30^\circ=\tfrac{\pi}{6}$ . We need the second quadrant (where $\tan$ is negative and $\tfrac{\pi}{2}<x<\pi$ ): take $\pi-\tfrac{\pi}{6}$ .

Answer

x=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}

Where the marks go

Common mistake

When the question says exact value or surd form, never round.

\tfrac{\sqrt3}{2}

is the answer;

0.866

earns nothing.

Common mistake

The basic angle is always measured to the

x

-axis, not the

y

-axis. For

120^\circ

that is

180-120=60^\circ

, not

30^\circ

Now you try

Without a calculator, find the exact value of $\cos\tfrac{7\pi}{6}$ . (9709-style)

Your turn— tap to reveal the worked answer (third quadrant, basic angle π/6)

$\tfrac{7\pi}{6}$ is in the third quadrant. The basic angle is $\tfrac{7\pi}{6}-\pi=\tfrac{\pi}{6}$ . There only Tan is positive, so the cosine is negative:

Answer

\cos\tfrac{7\pi}{6}=-\cos\tfrac{\pi}{6}=-\tfrac{\sqrt3}{2}