M5.3.2 Chapter 2

The unit circle and any angle

How can I find sin 150° or cos 225° when the angle does not fit in a right angled triangle?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A sin 150° is the same as sin 30°.
You picked A. Read on. At the end, the chapter comes back to your guess.
B sin 150° does not exist, because a triangle cannot have an angle of 150° and a right angle.
You picked B. Read on. At the end, the chapter comes back to your guess.
C sin 150° is negative, because 150° is more than 90°.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use the unit circle to find sin and cos of any angle
Find the sign of sin, cos and tan in each quarter

In a right angled triangle, sin, cos and tan are ratios of sides. That works only for angles under 90°, because a triangle with a right angle cannot hold another angle that big. But a wheel, a swing or a sound wave turns through any angle. We need sin and cos for every angle.

The unit circle is a circle with radius 1 and its centre at the origin. Start at the point (1, 0). Turn round the circle through an angle θ, anticlockwise for positive θ. Call the point you reach P.

The radius OP has length 1. Its end P has x coordinate cos θ and y coordinate sin θ.

For an acute angle, the radius OP is the hypotenuse of a right angled triangle with hypotenuse 1. The side next to θ is the x coordinate of P and the side opposite θ is its y coordinate. So cos θ = adjacent ÷ 1 = x, and sin θ = opposite ÷ 1 = y.

This gives a definition that works for every angle. Definition. For any angle θ, the point P reached on the unit circle has coordinates P = (cos θ, sin θ). So cos θ is the x coordinate and sin θ is the y coordinate. And tan θ = sin θ ÷ cos θ, which is the gradient of OP, whenever cos θ is not 0.

Because x and y lie between −1 and 1, we always have −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1. A negative angle turns clockwise. After a full turn of 360° (2π radians) the point is back where it started, so sin(θ + 360°) = sin θ and cos(θ + 360°) = cos θ.

The axes divide the plane into four quadrants, numbered anticlockwise from the top right. In each quadrant x and y have a fixed sign, so sin θ and cos θ have a fixed sign. This gives the signs of all three ratios.

The signs of sin, cos and tan in each quadrant. The sign of tan is the sign of sin divided by the sign of cos.

Each quadrant names what is positive there. Start in the fourth quadrant and go anticlockwise: Cos, All, Sin, Tan. The first letters spell CAST.

Angles in the other quadrants give the same numbers as acute angles, with a sign. The acute angle between the radius OP and the x axis is the reference angle. It is θ in the first quadrant, 180° − θ in the second, θ − 180° in the third and 360° − θ in the fourth.

The ray at 150° makes 30° with the negative x axis. The ray at 225° makes 45° with it. These acute angles are the reference angles.

The point at 150° is the mirror image of the point at 30° in the y axis. It has the same height, so sin 150° = sin 30°. Its x coordinate has the opposite sign, so cos 150° = −cos 30°. Reflection gives every other quadrant the same way.

P at 30° and Q at 150° are reflections in the y axis. They have the same y coordinate, so sin 150° = sin 30°. The x coordinates are opposites.

Rule. To find sin, cos or tan of any angle, follow four steps. First find the quadrant. Then find the reference angle. Then take the exact or calculator value for that angle. Last, give it the sign for the quadrant. For angles on the axes, read the point itself: it is (1, 0) at 0°, (0, 1) at 90°, (−1, 0) at 180° and (0, −1) at 270°.

Reflections and turns give results worth learning. For every x: sin(−x) = −sin x, cos(−x) = cos x and tan(−x) = −tan x where tan x is defined. Also sin(π − x) = sin x and cos(π − x) = −cos x. And sin(π + x) = −sin x and cos(π + x) = −cos x. Last, sin(π/2 − x) = cos x and cos(π/2 − x) = sin x. In degrees, replace π by 180°.

Each is a mirror image or a half turn of the point at x on the unit circle. A full turn, 2π or 360°, brings the point back, so sin(x + 2nπ) = sin x and cos(x + 2nπ) = cos x for every integer n. For an angle of more than one turn, or a negative angle, take away or add whole turns first.

The exact values for 30°, 45° and 60° come from two triangles you know: half of an equilateral triangle, and half of a square.

Half of an equilateral triangle with side 2. So sin 30° = 1/2, cos 30° = √3/2 and tan 30° = 1/√3 = √3/3.

Half of a square with side 1. So sin 45° = 1/√2 = √2/2, cos 45° = √2/2 and tan 45° = 1.

The exact values to learn. At 90° the point is (0, 1), so cos 90° = 0 and tan 90° = 1/0 has no value.

Because tan θ = sin θ ÷ cos θ, tan θ is not defined when cos θ = 0. That happens at 90°, 270°, and every angle 180° apart from these. In radians these are π/2, 3π/2, and so on. Near these angles tan θ becomes very large, positive on one side and negative on the other.

Worked example

Find the exact values of sin 150°, cos 150° and tan 150°.

150° lies between 90° and 180°, so it is in the second quadrant.
The reference angle is 180° − 150° = 30°.
We know sin 30° = 1/2, cos 30° = √3/2 and tan 30° = √3/3.
In the second quadrant sin is positive, cos is negative and tan is negative.
So sin 150° = 1/2, cos 150° = −√3/2 and tan 150° = −√3/3.
Check: tan 150° = sin 150° ÷ cos 150° = (1/2) ÷ (−√3/2) = −1/√3 = −√3/3.

Answer: sin 150° = 1/2, cos 150° = −√3/2, tan 150° = −√3/3.

Worked example

Given that cos θ = −5/13 and that 180° < θ < 270°, find sin θ and tan θ.

The point P = (cos θ, sin θ) lies on the unit circle, so x² + y² = 1.
Here x = −5/13, so x² = 25/169 and y² = 1 − 25/169.
Write 1 as 169/169. Then the top is 169 − 25 = 144.
So y² = 144/169. The square roots are 13 on the bottom and 12 on the top.
So y = 12/13 or y = −12/13, because both have a square of 144/169.
θ is in the third quadrant, where sin is negative. So sin θ = −12/13.
Then tan θ = sin θ ÷ cos θ = (−12/13) ÷ (−5/13) = 12/5.

Answer: sin θ = −12/13 and tan θ = 12/5.

The point P is in the third quadrant, so both of its coordinates are negative: cos θ = −5/13 and sin θ = −12/13.

Worked example

Find the exact values of cos(7π/6), sin(−π/3) and tan(5π/4).

Change each to degrees if that helps. 7π/6 = 210°, −π/3 = −60° and 5π/4 = 225°.
First cos 210°. It is in the third quadrant. The reference angle is 210° − 180° = 30°.
cos 30° = √3/2, and cos is negative in the third quadrant. So cos(7π/6) = −√3/2.
Next sin(−60°). A turn of 60° clockwise reaches the fourth quadrant, where sin is negative.
The reference angle is 60° and sin 60° = √3/2. So sin(−π/3) = −√3/2.
Last tan 225°. It is in the third quadrant, with reference angle 225° − 180° = 45°.
tan 45° = 1, and tan is positive in the third quadrant. So tan(5π/4) = 1.

Answer: cos(7π/6) = −√3/2, sin(−π/3) = −√3/2 and tan(5π/4) = 1.

Worked example

Find the exact values of cos(25π/6) and sin(−750°).

25π/6 is more than a full turn. Write 25π/6 = 4π + π/6, where 4π is two full turns.
So cos(25π/6) = cos(π/6) = √3/2.
For −750°, add two full turns, 720°. Then −750° + 720° = −30°.
So sin(−750°) = sin(−30°) = −sin 30° = −1/2.

Answer: cos(25π/6) = √3/2 and sin(−750°) = −1/2

See it three ways

With things

A child on a wheel that turns steadily goes up and down. At the start the child is level with the centre. After a quarter turn the child is at the top, after a half turn level again, and after three quarters at the bottom.

As a picture

In words

sin θ is the height of the point above the x axis, and cos θ is how far it is to the right of the y axis. Both are negative when the point is below or to the left.

In symbols

P = (cos θ, sin θ). At θ = 150°, P = (−√3/2, 1/2), so sin 150° = 1/2 and cos 150° = −√3/2.

Try it

Measure sin and cos on a big circle

You need: paper, a ruler, a protractor, a pencil.

Draw an x axis and a y axis that cross in the middle of the paper. Mark 10 cm from the crossing on each side. Let 10 cm stand for 1.
Draw a circle of radius 10 cm with its centre at the crossing.
With the protractor, mark 150° from the right hand end of the x axis, turning anticlockwise. Draw the radius to that mark.
From the end of the radius, measure down to the x axis, and measure across to the y axis. Divide each length by 10. The one across is negative, because it is to the left.

What you will see: The height is about 0.5 and the distance across is about −0.87. These are sin 150° and cos 150°. Try 30° and see the same height.

If you cannot try it, this is what happens.

A wrong idea many people have

sin 150° = −1/2, because 150° is bigger than 90° and sin goes negative there.

Why it feels right: The cosine went negative after 90°, and the sine seems to follow it.

The test that breaks it: The point at 150° is above the x axis, so its height sin 150° is positive. The graph of the point turning round the circle only goes below the x axis after 180°.

What is true: sin is positive in the first and second quadrants, from 0° to 180°. Use the quadrant, not whether the angle is above 90°.

A wrong idea many people have

sin 150° = sin 60°, because 150° = 90° + 60°.

Why it feels right: You may subtract 90° as a habit from a right angled triangle, which has an angle of 90°.

The test that breaks it: Look at the point at 150° and its mirror at 30°. Both are at height 1/2. But sin 60° = √3/2, which is a different height.

What is true: The reference angle is the acute angle with the x axis, not with the y axis. In the second quadrant it is 180° − θ. So the reference angle is 30°.

A wrong idea many people have

tan 90° = 1 ÷ 0 = 0.

Why it feels right: Dividing 1 by a small number gives a big number, and dividing by nothing seems to give nothing.

The test that breaks it: Try 1 ÷ 0.1 = 10, 1 ÷ 0.01 = 100, 1 ÷ 0.001 = 1000. The smaller the bottom, the bigger the answer. It does not settle at 0.

What is true: Division by 0 has no value, so tan 90° is not defined. The same is true at 270° and at every angle 180° from these.

Key words

unit circle: the circle with radius 1 and its centre at the origin.
quadrant: one of the four regions into which the axes cut the plane, numbered 1 to 4 anticlockwise from the top right.
reference angle: the acute angle between the radius OP and the x axis.

Back to your guess

The answer is A: sin 150° is the same as sin 30°.

sin 150° = sin 30° = 1/2. A triangle cannot show it, but a circle can. We turn through 150° round a circle of radius 1 and read the height of the point we reach. That height is the same as at 30°.

Keep it

Copy this into your exercise book.

P = (cos θ, sin θ) on the unit circle. Signs: first All, second Sin, third Tan, fourth Cos. Size from the reference angle. tan θ = sin θ ÷ cos θ, none when cos θ = 0.

Draw: A circle with axes, four quadrants marked A, S, T, C, and one radius with its reference angle.