Special Angles

Most trig values are messy decimals, like $\sin 37^\circ \approx 0.6018$. But a few angles, $30^\circ$, $45^\circ$, and $60^\circ$, have exact values you can find without a calculator. They come up constantly, so it’s worth knowing where they come from and being able to rebuild them quickly. All angles on this page are in degrees.

Key ideas

The primary trig ratios (a reminder)

For an acute angle $\theta$ in a right triangle:

$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$

(“SOH CAH TOA” is a handy way to remember them.)

Two special triangles

• The $45^\circ$–$45^\circ$–$90^\circ$ triangle is half of a square with side $1$. Both legs are $1$, and by the Pythagorean theorem the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
• The $30^\circ$–$60^\circ$–$90^\circ$ triangle is half of an equilateral triangle with side $2$. The hypotenuse is $2$, the short leg (opposite $30^\circ$) is $1$, and the long leg is $\sqrt{2^2 - 1^2} = \sqrt{3}$.

The exact values

Reading the ratios from the triangles:

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	$0$	$\tfrac{1}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{\sqrt{3}}{2}$	$1$
$\cos\theta$	$1$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{1}{2}$	$0$
$\tan\theta$	$0$	$\tfrac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	undefined

• $\sin 45^\circ = \tfrac{1}{\sqrt{2}}$, which is the same as $\tfrac{\sqrt{2}}{2}$ (multiply the top and bottom by $\sqrt{2}$). Likewise $\tan 30^\circ = \tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3}$.
• $\tan 90^\circ = \dfrac{\sin 90^\circ}{\cos 90^\circ} = \dfrac{1}{0}$, which is undefined.

A memory trick for the sine row: $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$. The cosine row is the same list backwards.

Worked examples

Example 1: Combining exact values

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

Solution.

$$\sin 60^\circ + \cos 30^\circ = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}$$

Example 2: A product

Find the exact value of $2\sin 45^\circ \cos 45^\circ$.

Solution.

$$2\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = 2\left(\frac{2}{4}\right) = 1$$

Example 3: Squares of sine and cosine

Find the exact value of $(\sin 30^\circ)^2 + (\cos 30^\circ)^2$.

Solution.

$$\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$$

(This isn’t a coincidence. You’ll see why it always equals $1$ in trig identities.)

Example 4: Exact side lengths

A $6$ m ladder leans against a wall, making a $60^\circ$ angle with the ground. How high up the wall does it reach? Give an exact answer and a decimal.

Solution. The height is opposite the $60^\circ$ angle, and the ladder is the hypotenuse:

$$h = 6\sin 60^\circ = 6\left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3} \approx 5.20$$

The ladder reaches $3\sqrt{3}$ m, about $5.20$ m, up the wall.

Common mistakes

Mixing up $\sin 30^\circ$ and $\sin 60^\circ$. In the $30^\circ$–$60^\circ$–$90^\circ$ triangle, the side opposite $30^\circ$ is the shortest side, $1$. So $\sin 30^\circ = \tfrac{1}{2}$, and $\sin 60^\circ = \tfrac{\sqrt{3}}{2}$.

Using the wrong side as the hypotenuse. The hypotenuse is always opposite the right angle, and it’s the longest side: $\sqrt{2}$ and $2$ in the special triangles.

Writing $\tan 90^\circ = 0$. It’s undefined, because it would mean dividing by $\cos 90^\circ = 0$.

Squaring incorrectly. $\left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{3}{4}$, not $\tfrac{\sqrt{3}}{4}$ or $\tfrac{9}{4}$.

Using radian mode. These values are for degrees. If your calculator gives $\sin 30 \approx -0.988$, it’s in radian mode.

Practice

1. (Warm-up) Give the exact values of $\sin 30^\circ$, $\cos 45^\circ$, and $\tan 60^\circ$.

Solution

$\tfrac{1}{2}$, $\tfrac{\sqrt{2}}{2}$, and $\sqrt{3}$.

2. (Warm-up) Find the exact value of $\cos 60^\circ + \sin 30^\circ$.

Solution

$\tfrac{1}{2} + \tfrac{1}{2} = 1$

3. (Warm-up) Find the exact value of $\tan 45^\circ \times \sin 90^\circ$.

Solution

$1 \times 1 = 1$

4. (Core) Find the exact value of $\sin 45^\circ \cos 45^\circ$.

Solution

$$\frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = \frac{2}{4} = \frac{1}{2}$$

5. (Core) Find the exact value of $\tan 30^\circ \tan 60^\circ$.

Solution

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

6. (Core) Find the exact value of $4\sin^2 60^\circ - 3$. (Here $\sin^2 60^\circ$ means $(\sin 60^\circ)^2$.)

Solution

$$4\left(\frac{\sqrt{3}}{2}\right)^2 - 3 = 4\left(\frac{3}{4}\right) - 3 = 3 - 3 = 0$$

7. (Core) A right triangle has a $45^\circ$ angle and a hypotenuse of $8$ cm. Find the exact lengths of the legs.

Solution

Each leg is $8\sin 45^\circ = 8\left(\tfrac{\sqrt{2}}{2}\right) = 4\sqrt{2}$ cm.

8. (Core) A kite string is $30$ m long and makes a $60^\circ$ angle with the ground. How high is the kite? Give an exact answer and a decimal to two places.

Solution

$$h = 30\sin 60^\circ = 30\left(\frac{\sqrt{3}}{2}\right) = 15\sqrt{3} \approx 25.98$$

The kite is $15\sqrt{3}$ m, about $25.98$ m, high.

9. (Challenge) Show that $1 + \tan^2 45^\circ = \dfrac{1}{\cos^2 45^\circ}$.

Solution

Left side: $1 + 1^2 = 2$.

Right side: $\dfrac{1}{\left(\frac{\sqrt{2}}{2}\right)^2} = \dfrac{1}{\frac{1}{2}} = 2$.

Both sides equal $2$. ✓

10. (Challenge) An equilateral triangle has an altitude (height) of $6$ cm. Find its exact perimeter.

Solution

The altitude splits it into two $30^\circ$–$60^\circ$–$90^\circ$ triangles. The altitude is opposite the $60^\circ$ angle, and the side $s$ of the equilateral triangle is the hypotenuse:

$$\sin 60^\circ = \frac{6}{s} \quad\Rightarrow\quad s = \frac{6}{\frac{\sqrt{3}}{2}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$$

The perimeter is $3 \times 4\sqrt{3} = 12\sqrt{3}$ cm.