Exact values of sin, cos and tan

By the end of this lesson

By the end you can write down the exact values of $\sin$ , $\cos$ and $\tan$ of $30^\circ$ , $45^\circ$ and $60^\circ$ (and $\tfrac{\pi}{6}$ , $\tfrac{\pi}{4}$ , $\tfrac{\pi}{3}$ ), and of related angles such as $210^\circ$ or $\tfrac{5\pi}{4}$ , without a calculator.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$\theta$	an angle
exact	written with fractions and surds like $\sqrt3$ , never a rounded decimal

Two triangles you can draw in 10 seconds

No memorising a wall of surds. Everything comes from two triangles:

Half an equilateral triangle.

Take an equilateral triangle with every side 2. All its angles are $60^\circ$ .
Cut it down the middle. You get a right-angled triangle with hypotenuse 2 and base 1.
The top angle is halved, to $30^\circ$ . The bottom angle stays $60^\circ$ .
The height, by Pythagoras: $\sqrt{2^2 - 1^2} = \sqrt{4 - 1} = \sqrt3$ .

Half a square.

Take a square with sides 1. Cut it along a diagonal.
You get a right-angled triangle with two sides of 1 and two $45^\circ$ angles.
The hypotenuse, by Pythagoras: $\sqrt{1^2 + 1^2} = \sqrt2$ .

Then it's just SOH CAH TOA. For example, in the first triangle the side opposite $30^\circ$ is 1 and the hypotenuse is 2:

$\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac12$

The table

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
$30^\circ = \tfrac{\pi}{6}$	$\tfrac12$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{\sqrt3} = \tfrac{\sqrt3}{3}$
$45^\circ = \tfrac{\pi}{4}$	$\tfrac{1}{\sqrt2} = \tfrac{\sqrt2}{2}$	$\tfrac{1}{\sqrt2} = \tfrac{\sqrt2}{2}$	$1$
$60^\circ = \tfrac{\pi}{3}$	$\tfrac{\sqrt3}{2}$	$\tfrac12$	$\sqrt3$

Why two forms like $\tfrac{1}{\sqrt3} = \tfrac{\sqrt3}{3}$ ? Multiply top and bottom by $\sqrt3$ :

$\frac{1 \times \sqrt3}{\sqrt3 \times \sqrt3} = \frac{\sqrt3}{3}$

Same number, no root on the bottom. Either form is exact.

Also handy, from the unit circle:

$\sin 0 = 0$ and $\cos 0 = 1$
$\sin 90^\circ = 1$ and $\cos 90^\circ = 0$
$\tan 90^\circ$ is undefined (it would mean dividing by $\cos 90^\circ = 0$ )

The memory hack

Read the sine column as $\tfrac{\sqrt1}{2}, \tfrac{\sqrt2}{2}, \tfrac{\sqrt3}{2}$ for $30^\circ, 45^\circ, 60^\circ$ . Cosine is the same list backwards. Then $\tan = \tfrac{\sin}{\cos}$ .

Sanity check: as an acute angle grows, sine grows and cosine shrinks. If your $\sin 60^\circ$ is smaller than your $\sin 30^\circ$ , something's swapped.

Quick check

Multiple choiceQuestion 1 · Recall

Answer the question above to continue.