Math Core

Most trig values, like $\sin 37^\circ$ , are messy decimals that you need a calculator for. But three angles, $30^\circ$ , $45^\circ$ and $60^\circ$ , have exact values you can find with nothing but the Pythagorean theorem. They show up constantly in trigonometry, so it pays to know them cold.

The 45°-45°-90° triangle

Cut a square along its diagonal. Each half is an isosceles right triangle with angles $45^\circ$ , $45^\circ$ and $90^\circ$ . If each leg has length $s$ , the Pythagorean theorem gives the hypotenuse:

\sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}.

A 45°-45°-90° triangle: two equal legs, hypotenuse √2 times a leg.

The 30°-60°-90° triangle

Start with an equilateral triangle with side $2s$ and draw an altitude. The altitude splits it into two congruent right triangles, each with angles $30^\circ$ , $60^\circ$ and $90^\circ$ . In each half:

the hypotenuse is a full side of the equilateral triangle: $2s$ ;
the short leg is half of the base: $s$ ;
the long leg (the altitude) is $\sqrt{(2s)^2 - s^2} = \sqrt{3s^2} = s\sqrt{3}$ .

A 30°-60°-90° triangle: short leg s across from 30°, long leg s√3 across from 60°, hypotenuse 2s.

The short leg is across from the smallest angle ( $30^\circ$ ), and the long leg is across from $60^\circ$ .

Side ratios of the special right triangles

triangle	sides
$45^\circ$ - $45^\circ$ - $90^\circ$	leg : leg : hypotenuse $= 1 : 1 : \sqrt{2}$
$30^\circ$ - $60^\circ$ - $90^\circ$	short leg : long leg : hypotenuse $= 1 : \sqrt{3} : 2$

Worked example: Finding the legs of a 45°-45°-90° triangle

The hypotenuse of a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle is $10$ . How long is each leg?

The hypotenuse is $s\sqrt{2}$ , so $s\sqrt{2} = 10$ and

s = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}.

Each leg is $5\sqrt{2} \approx 7.07$ .

Worked example: Starting from the long leg

In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the long leg is $9$ . Find the short leg and the hypotenuse.

The long leg is $s\sqrt{3}$ , so

s = \frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}.

The short leg is $3\sqrt{3}$ and the hypotenuse is $2s = 6\sqrt{3}$ .

Tip

When you know the long leg or the hypotenuse, always find the short leg $s$ first. Every other side is a simple multiple of it.

Exact trig values

Now read the six ratios straight off the two triangles. Using $s = 1$ keeps the arithmetic simple.

From the $45^\circ$ - $45^\circ$ - $90^\circ$ triangle (legs $1$ , hypotenuse $\sqrt{2}$ ):

\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \tan 45^\circ = \frac{1}{1} = 1.

From the $30^\circ$ - $60^\circ$ - $90^\circ$ triangle (legs $1$ and $\sqrt{3}$ , hypotenuse $2$ ), stand at the $30^\circ$ angle: the opposite side is $1$ and the adjacent side is $\sqrt{3}$ . At the $60^\circ$ angle they switch.

\sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3}

The reciprocal ratios follow by flipping. Put together:

Exact values at 30°, 45° and 60°

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$	$\csc\theta$	$\sec\theta$	$\cot\theta$
$30^\circ$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$	$2$	$\frac{2\sqrt{3}}{3}$	$\sqrt{3}$
$45^\circ$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$	$\sqrt{2}$	$\sqrt{2}$	$1$
$60^\circ$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$	$\frac{2\sqrt{3}}{3}$	$2$	$\frac{\sqrt{3}}{3}$

You don't need to memorize the table as a wall of numbers. Notice the patterns:

The sine column climbs $\dfrac{1}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}$ (think $\dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}$ ), and cosine is the same list backwards. That is the cofunction identity: $\cos 30^\circ = \sin 60^\circ$ .
Tangent is sine divided by cosine, and $\tan 30^\circ$ and $\tan 60^\circ$ are reciprocals, just like $30^\circ$ and $60^\circ$ are complements.
If you forget a value, sketch the triangle with sides $1$ , $\sqrt{3}$ , $2$ or $1$ , $1$ , $\sqrt{2}$ and read it off.

Common mistake

The most common slip is mixing up which leg of the $30^\circ$ - $60^\circ$ - $90^\circ$ triangle is which. The shorter leg is across from the smaller angle. So $\sin 30^\circ = \dfrac{1}{2}$ (short leg over hypotenuse), not $\dfrac{\sqrt{3}}{2}$ . A quick sanity check: $30^\circ$ is a small angle, so its sine should be the smaller of the two values.

Worked example: Evaluating an expression exactly

Find the exact value of $\sin 60^\circ \cos 30^\circ + \tan^2 45^\circ$ .

(Here $\tan^2 45^\circ$ means $(\tan 45^\circ)^2$ .)

\sin 60^\circ \cos 30^\circ + \tan^2 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + 1^2 = \frac{3}{4} + 1 = \frac{7}{4}.

Worked example: The height of an equilateral triangle

An equilateral triangle has side length $8$ . Find its height and its area exactly.

The altitude cuts the triangle into two $30^\circ$ - $60^\circ$ - $90^\circ$ triangles with hypotenuse $8$ . So the short leg is $s = 4$ and the height (the long leg) is $4\sqrt{3}$ . You can also get it with trig: $\text{height} = 8 \sin 60^\circ = 8 \cdot \dfrac{\sqrt{3}}{2} = 4\sqrt{3}$ .

\text{Area} = \frac{1}{2}(8)(4\sqrt{3}) = 16\sqrt{3}.

Practice

Practice 1

A $45^\circ$ - $45^\circ$ - $90^\circ$ triangle has legs of length $7$ . What is the exact length of the hypotenuse?