# Trig values of 36 degrees

## Statement

The exact value of the trigonometry functions of $36 \degree$ are:

$$\sin(36 \degree) = \tfrac{1}{4} \sqrt{10 - 2\sqrt5}$$

$$\cos(36 \degree) = \tfrac{1}{4} \sqrt{6 + 2\sqrt5}$$

$$\tan(36 \degree) = \sqrt{5 - 2\sqrt5}$$

## Sine

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

When substituting $18 \degree$ for $\alpha$ and $\beta$, you get:

$$\sin(36 \degree) = \sin(18 \degree + 18 \degree) = \sin(18 \degree) \cos(18 \degree) + \cos(18 \degree) \sin(18 \degree) = 2 \sin(18 \degree) \cos(18 \degree)$$

$$\sin(18 \degree) = \tfrac{1}{4} \sqrt{6 - 2\sqrt5}$$

$$\cos(18 \degree) = \tfrac{1}{4} \sqrt{10 + 2\sqrt5}$$

Now substitute these values, and simplify:

\begin{aligned}\sin(36 \degree) &= 2 \sin(18 \degree) \cos(18 \degree) \newline&= 2 * \tfrac{1}{4} \sqrt{6 - 2\sqrt5} * \tfrac{1}{4} \sqrt{10 + 2\sqrt5} \newline&= \tfrac{1}{8} \sqrt{(6 - 2\sqrt5)(10 + 2\sqrt5)} \newline&= \tfrac{1}{8} \sqrt{60 + 12\sqrt5 - 20\sqrt5 - 20} \newline&= \tfrac{1}{8} \sqrt{40 - 8\sqrt5} \newline&= \tfrac{1}{8} \sqrt{4(10 - 2\sqrt5)} \newline&= \tfrac{1}{4} \sqrt{10 - 2\sqrt5}\end{aligned}

## Cosine

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

When substituting $18 \degree$ for $\alpha$ and $\beta$, you get:

$$\cos(36 \degree) = \cos(18 \degree + 18 \degree) = \cos(18 \degree) \cos(18 \degree) - \sin(18 \degree) \sin(18 \degree) = \cos(18 \degree)^2 - \sin(18 \degree)^2$$

$$\sin(18 \degree) = \tfrac{1}{4} \sqrt{6 - 2\sqrt5}$$

$$\cos(18 \degree) = \tfrac{1}{4} \sqrt{10 + 2\sqrt5}$$

Now substitute these values, and simplify:

\begin{aligned}\cos(36 \degree) &= \cos(18 \degree)^2 - \sin(18 \degree)^2 \newline&= \left(\tfrac{1}{4} \sqrt{10 + 2\sqrt5}\right)^2 - \left(\tfrac{1}{4} \sqrt{6 - 2\sqrt5}\right)^2 \newline&= \tfrac{1}{16} \left(10 + 2\sqrt5 \right) - \tfrac{1}{16} \left(6 - 2\sqrt5 \right) \newline&= \tfrac{10}{16} + \tfrac{2}{16} \sqrt5 - \tfrac{6}{16} + \tfrac{2}{16} \sqrt5 \newline&= \tfrac{1}{4} + \tfrac{1}{4} \sqrt5 \newline&= \tfrac{1}{4} \left(1 + \sqrt5 \right) \newline&= \tfrac{1}{4} \sqrt{\left(1 + \sqrt5 \right)^2} \newline&= \tfrac{1}{4} \sqrt{6 + 2\sqrt5}\end{aligned}

## Tangent

Use that tangent is sine divided by cosine and simplify.

\begin{aligned}\tan(36 \degree) &= \frac{\sin(36 \degree)}{\cos(36 \degree)} \newline&= \frac{\tfrac{1}{4} \sqrt{10 - 2\sqrt5}}{\tfrac{1}{4} \sqrt{6 + 2\sqrt5}} \newline&= \sqrt{\frac{10 - 2\sqrt5}{6 + 2\sqrt5}} \newline&= \sqrt{\frac{10 - 2\sqrt5}{6 + 2\sqrt5} * \frac{6 - 2\sqrt5}{6 - 2\sqrt5}} \newline&= \sqrt{\frac{60 - 20\sqrt5 - 12\sqrt5 + 20}{36 - 20}} \newline&= \sqrt{\frac{80 - 32\sqrt5}{16}} \newline&= \sqrt{\frac{16 \left(5 - 2\sqrt5 \right)}{16}} \newline&= \sqrt{5 - 2\sqrt5}\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 54 degrees

This proof shows the exact value of sin(54), cos(54) and tan(54)