# Trig values of 15 degrees

## Statement

The exact value of the trigonometry functions of $15 \degree$ are:

$$\sin(15 \degree) = \tfrac{1}{4} \sqrt{8 - 4\sqrt3}$$

$$\cos(15 \degree) = \tfrac{1}{4} \sqrt{8 + 4\sqrt3}$$

$$\tan(15 \degree) = \sqrt{7 - 4\sqrt3}$$

## Sine

$$\sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \sin(\beta) \cos(\alpha)$$

When substituting $45 \degree$ for $\alpha$ and $30 \degree$ for $\beta$, you get:

$$\sin(15 \degree) = \sin(45 \degree - 30 \degree) = \sin(45 \degree) \cos(30 \degree) - \sin(30 \degree) \cos(45 \degree)$$

$$\sin(30 \degree) = \tfrac{1}{2}$$

$$\cos(30 \degree) = \tfrac{1}{2} \sqrt3$$

$$\sin(45 \degree) = \tfrac{1}{2} \sqrt2$$

$$\cos(45 \degree) = \tfrac{1}{2} \sqrt2$$

Now substitute these values, and simplify:

\begin{aligned}\sin(15 \degree) &= \sin(45 \degree) \cos(30 \degree) - \sin(30 \degree) \cos(45 \degree) \newline&= \tfrac12 \sqrt2 \cdot \tfrac12 \sqrt3 - \tfrac12 \cdot \tfrac12 \sqrt2 \newline&= \tfrac14 \sqrt6 - \tfrac14 \sqrt2 \newline&= \tfrac14 \sqrt{\left(\sqrt6 - \sqrt2\right)^2} \newline&= \tfrac14 \sqrt{6 - 2\sqrt{12} + 2} \newline&= \tfrac14 \sqrt{8 - 4\sqrt3}\end{aligned}

## Cosine

$$\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$$

When substituting $45 \degree$ for $\alpha$ and $30 \degree$ for $\beta$, you get:

$$\cos(15 \degree) = \cos(45 \degree - 30 \degree) = \cos(45 \degree) \cos(30 \degree) + \sin(45 \degree) \sin(30 \degree)$$

$$\sin(30 \degree) = \tfrac{1}{2}$$

$$\cos(30 \degree) = \tfrac{1}{2} \sqrt3$$

$$\sin(45 \degree) = \tfrac{1}{2} \sqrt2$$

$$\cos(45 \degree) = \tfrac{1}{2} \sqrt2$$

Now substitute these values, and simplify:

\begin{aligned}\cos(15 \degree) &= \cos(45 \degree) \cos(30 \degree) + \sin(45 \degree) \sin(30 \degree) \newline&= \tfrac12 \sqrt2 \cdot \tfrac12 \sqrt3 + \tfrac12 \sqrt2 \cdot \tfrac12 \newline&= \tfrac14 \sqrt6 + \tfrac14 \sqrt2 \newline&= \tfrac14 \sqrt{\left(\sqrt6 + \sqrt2\right)^2} \newline&= \tfrac14 \sqrt{6 + 2\sqrt{12} + 2} \newline&= \tfrac14 \sqrt{8 + 4\sqrt3}\end{aligned}

## Tangent

Use that tangent is sine divided by cosine and simplify.

\begin{aligned}\tan(15 \degree) &= \frac{\sin(15 \degree)}{\cos(15 \degree)} \newline&= \frac{\tfrac14 \sqrt{8 - 4\sqrt3}}{\tfrac14 \sqrt{8 + 4\sqrt3}} \newline&= \sqrt{\frac{8 - 4\sqrt3}{8 + 4\sqrt3}} \newline&= \sqrt{\frac{8 - 4\sqrt3}{8 + 4\sqrt3} \cdot \frac{8 - 4\sqrt3}{8 - 4\sqrt3}} \newline&= \sqrt{\frac{64 - 64\sqrt3 + 48}{64 - 48}} \newline&= \sqrt{\frac{16(4 - 4\sqrt3 + 3)}{16}} \newline&= \sqrt{7 - 4\sqrt3} \newline\end{aligned}

## Proofs building upon this proof

### Exact values of the trig functions

This table shows all exact values of sine, cosine and tangent.

### Trig values of 3 degrees

This proof shows the exact value of sin(3), cos(3) and tan(3).