Exact values and angles of any size

Paper 1 expects you to know the sine, cosine and tangent of $30^\circ$, $45^\circ$ and $60^\circ$ exactly, and to work out related values such as $\cos 150^\circ = -\tfrac{1}{2}\sqrt{3}$ or $\sin\tfrac{3\pi}{4} = \tfrac{1}{2}\sqrt{2}$ without a calculator. These exact values appear in "show that" questions, in exact areas of sectors and triangles, and in every trigonometric equation with a neat answer. This note explains where the values come from and how the unit circle extends them to angles of any size, positive or negative.

Where the exact values come from

Two triangles give every value you need.

Half an equilateral triangle. An equilateral triangle of side $2$ has all angles $60^\circ$. Cutting it in half down its line of symmetry gives a right-angled triangle with hypotenuse $2$, short side $1$, angles $30^\circ$ and $60^\circ$, and (by Pythagoras) third side $\sqrt{4 - 1} = \sqrt{3}$.

Half a square. A square of side $1$ cut along a diagonal gives a right-angled isosceles triangle with angles $45^\circ$, sides $1$ and $1$, and hypotenuse $\sqrt{2}$.

Reading off opposite, adjacent and hypotenuse gives the table.

A memory aid: the sines of $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ are $\tfrac{\sqrt{0}}{2}, \tfrac{\sqrt{1}}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}, \tfrac{\sqrt{4}}{2}$, and the cosines are the same list backwards. Then $\tan = \tfrac{\sin}{\cos}$. But the triangles are better: if you can draw them, you can never misremember a value.

Angles of any size

The unit circle

As in Graphs of sine, cosine and tangent, measure an angle $\theta$ anticlockwise from the positive $x$-axis, and let $P$ be the point at that angle on the circle of radius $1$. Then $P = (\cos\theta, \sin\theta)$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$.

• Negative angles are measured clockwise.
• Adding $360^\circ$ ($2\pi$) brings you back to the same point, so $\sin(\theta + 360^\circ) = \sin\theta$, and similarly for cosine. Tangent repeats every $180^\circ$.

Signs in the four quadrants

The signs of $\sin\theta$ and $\cos\theta$ are the signs of the $y$- and $x$-coordinates of $P$.

Quadrant	Angle (degrees)	Angle (radians)	Positive
first	$0^\circ$ to $90^\circ$	$0$ to $\tfrac{\pi}{2}$	all three
second	$90^\circ$ to $180^\circ$	$\tfrac{\pi}{2}$ to $\pi$	sine only
third	$180^\circ$ to $270^\circ$	$\pi$ to $\tfrac{3\pi}{2}$	tangent only
fourth	$270^\circ$ to $360^\circ$	$\tfrac{3\pi}{2}$ to $2\pi$	cosine only

Reading the positive functions anticlockwise from the fourth quadrant spells CAST, which is how many students remember it.

The points at $30^\circ$ and $150^\circ$ are mirror images in the $y$-axis: same height ($\sin 150^\circ = \sin 30^\circ = \tfrac{1}{2}$), opposite horizontal positions ($\cos 150^\circ = -\cos 30^\circ$).

The related acute angle

Every angle sits at some acute angle $\alpha$ to the $x$-axis, called its related angle (or reference angle). The values of $\sin$, $\cos$ and $\tan$ at $\theta$ are the values at $\alpha$, with a sign given by the quadrant.

These identities hold for every $\theta$, not only acute ones, and they can be read directly from the symmetries of the graphs.

Finding one ratio from another

If you know one of $\sin\theta$, $\cos\theta$ or $\tan\theta$ and which quadrant $\theta$ is in, you can find the other two exactly.

The identity $\sin^2\theta + \cos^2\theta = 1$ does the same job algebraically; see Trigonometric identities.

Worked examples

Practice questions

Answers

1. $210^\circ = 180^\circ + 30^\circ$ (third quadrant, cosine negative): $-\tfrac{\sqrt{3}}{2}$. $\tfrac{7\pi}{4} = 2\pi - \tfrac{\pi}{4}$ (fourth, sine negative): $-\tfrac{\sqrt{2}}{2}$. $\tfrac{5\pi}{6} = \pi - \tfrac{\pi}{6}$ (second, tangent negative): $-\tfrac{1}{\sqrt{3}}$.

2. $-300^\circ + 360^\circ = 60^\circ$: $\sin(-300^\circ) = \tfrac{\sqrt{3}}{2}$. $495^\circ - 360^\circ = 135^\circ$ (second quadrant): $\cos 495^\circ = -\tfrac{\sqrt{2}}{2}$. $\tfrac{11\pi}{3} - 4\pi = -\tfrac{\pi}{3}$: $\tan\tfrac{11\pi}{3} = -\sqrt{3}$.

3. Triangle $5$, $12$, $13$. Third quadrant: $\sin\theta = -\tfrac{12}{13}$, $\tan\theta = \tfrac{12}{5}$ (positive).

4. Triangle $3$, $4$, $5$. $\tan\theta > 0$ and reflex means the third quadrant ($180^\circ < \theta < 270^\circ$): $\sin\theta = -\tfrac{3}{5}$, $\cos\theta = -\tfrac{4}{5}$.

5. $\tfrac{\sqrt{3}}{2} \times \tfrac{\sqrt{3}}{2} + \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{3}{4} + \tfrac{1}{4} = 1$.

6. $\cos^2\tfrac{\pi}{6} - \sin^2\tfrac{\pi}{6} = \tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2} = \cos\tfrac{\pi}{3}$.

7. (a) $\cos x - \cos x = 0$. (b) $\tan x + \tan x = 2\tan x$. (c) $\sin x + \sin x - 2\sin x = 0$.

8. $QR^2 = 25 + 64 - 2(5)(8)\cos 60^\circ = 89 - 40 = 49$, so $QR = 7$ cm. Area $= \tfrac{1}{2}(5)(8)\sin 60^\circ = 20 \times \tfrac{\sqrt{3}}{2} = 10\sqrt{3}\ \text{cm}^2$.

9. Triangle with opposite $k$, hypotenuse $1$, adjacent $\sqrt{1 - k^2}$. Obtuse means second quadrant, so $\cos\theta = -\sqrt{1 - k^2}$ and $\tan\theta = -\dfrac{k}{\sqrt{1 - k^2}}$.

10. Triangle with adjacent $1$, hypotenuse $3$, opposite $\sqrt{9 - 1} = 2\sqrt{2}$. So $\sin\theta = \tfrac{2\sqrt{2}}{3}$ and $\tan\theta = 2\sqrt{2}$. Then $\sin(\pi + \theta) = -\sin\theta$ and $\cos(\pi - \theta) = -\cos\theta$, so the sum is $-\tfrac{2\sqrt{2}}{3} - \tfrac{1}{3} = -\dfrac{2\sqrt{2} + 1}{3}$.