# Trig values of 18 degrees

## Statement

The exact value of the trigonometry functions of $18 \degree$ are:

$$\sin(18 \degree) = \tfrac{1}{4} \sqrt{6 - 2 \sqrt{5}}$$

$$\cos(18 \degree) = \tfrac{1}{4} \sqrt{10 + 2 \sqrt{5}}$$

$$\tan(18 \degree) = \tfrac{1}{5} \sqrt{25 - 10 \sqrt{5}}$$

## Proof

Construct triangle $ABC$ with $\angle C = 36 \degree$ and $\angle A = \angle B = 72 \degree$. Then construct a bisector from $A$ to $D$ on $BC$, so that $\angle A_1 = \angle A_2 = 36 \degree$. Finally construct altitude $CE$ on $AB$, so that $\angle C_1 = \angle C_2 = 18 \degree$. See the image below.

Let $AC = m$. Because triangle $ABC$ is isosceles, $BC = AC = m$.

Let $AB = n$. Because triangle $ABD$ is isosceles, $AD = AB = n$. And because triangle $ADC$ is isosceles as well, $CD = AD = n$.

Note that $BE = \frac{1}{2} AB = \frac{1}{2} n$ and $BD = BC - CD = m - n$.

Now prove that triangle $ABC$ is similar to triangle $CAB$:

$$\begin{rcases} \angle A_1 = \angle C = 36 \degree \newline \angle B = \angle A = 72 \degree \newline \angle D_1 = \angle B = 72 \degree \end{rcases} \triangle ABD \sim \triangle CAB$$

From those sililar triangles, we get:

$$\frac{BD}{AB} = \frac{AB}{AC}$$

$$\frac{m - n}{n} = \frac{n}{m}$$

Now cross-multiply and divide each term by $m^2$.

$$m(m - n) = n * n$$

$$m^2 - mn = n^2$$

$$n^2 + mn - m^2 = 0$$

$$\frac{n^2}{m^2} + \frac{n}{m} - 1 = 0$$

$$\left(\frac{n}{m}\right)^2 + \left(\frac{n}{m}\right) - 1 = 0$$

Solve for $\dfrac{n}{m}$ using the quadratic formula (only positive values) and isolate $n$.

$$\frac{n}{m} = \frac{-1 + \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 + \sqrt{5}}{2}$$

$$n = \tfrac{1}{2} m (\sqrt{5} - 1)$$

Write $\sqrt{5} - 1$ as one square root.

$$n = \tfrac{1}{2} m \sqrt{(\sqrt{5} - 1)^2} = \tfrac{1}{2} m \sqrt{6 - 2 \sqrt{5}}$$

Find $CE$ using the Pythagorean theorem.

\begin{aligned}CE &= \sqrt{BC^2 - BE^2} \newline&= \sqrt{m^2 - (\tfrac{1}{2}n)^2} \newline&= \sqrt{m^2 - \left(\tfrac{1}{4}m \sqrt{6 - 2\sqrt{5}}\right)^2} \newline&= \sqrt{m^2 - \tfrac{1}{16} m^2 (6 - 2\sqrt{5})} \newline&= \sqrt{\tfrac{16}{16} m^2 - \tfrac{6}{16} m^2 + \tfrac{2\sqrt{5}}{16} m^2} \newline&= m * \frac{\sqrt{16 - 6 + 2 \sqrt{5}}}{\sqrt{16}} \newline&= \tfrac{1}{4} m \sqrt{10 + 2 \sqrt{5}}\end{aligned}

### Sine

From the triangle, note that $\sin(\angle C_1) = \dfrac{BE}{BC}$, so:

$$\sin(18 \degree) = \frac{\frac{1}{2}n}{m} = \frac{\frac{1}{2} * \tfrac{1}{2} m \sqrt{6 - 2 \sqrt{5}}}{m} = \tfrac{1}{4} \sqrt{6 - 2 \sqrt{5}}$$

### Cosine

From the triangle, note that $\cos(\angle C_1) = \dfrac{CE}{BC}$, so:

$$\cos(18 \degree) = \frac{\tfrac{1}{4} m \sqrt{10 + 2 \sqrt{5}}}{m} = \tfrac{1}{4} \sqrt{10 + 2 \sqrt{5}}$$

### Tangent

From the triangle, note that $\tan(\angle C_1) = \dfrac{BE}{CE}$, so:

\begin{aligned}\tan(18 \degree) &= \frac{\frac{1}{2}n}{\tfrac{1}{4} m \sqrt{10 + 2 \sqrt{5}}} \newline&= \frac{\frac{1}{2} * \tfrac{1}{2} m \sqrt{6 - 2 \sqrt{5}}}{\tfrac{1}{4} m \sqrt{10 + 2 \sqrt{5}}} \newline&= \frac{\sqrt{6 - 2\sqrt5}}{\sqrt{10 + 2\sqrt5}} \newline&= \sqrt{\frac{6 - 2\sqrt5}{10 + 2\sqrt5}} \newline&= \sqrt{\frac{6 - 2\sqrt5}{10 + 2\sqrt5} * \frac{10 - 2\sqrt5}{10 - 2\sqrt5}} \newline&= \sqrt{\frac{60 - 12\sqrt5 - 20\sqrt5 + 20}{100 - 20}} \newline&= \sqrt{\frac{80 - 32\sqrt5}{80}} \newline&= \sqrt{\frac{5 - 2\sqrt5}{5}} \newline&= \frac{\sqrt{5 - 2\sqrt5}}{\sqrt5} \newline&= \tfrac{1}{5} \sqrt5 \sqrt{5 - 2\sqrt5} \newline&= \tfrac{1}{5} \sqrt{25 - 10\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 3 degrees

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

### Trig values of 36 degrees

This proof shows the exact value of sin(36), cos(36) and tan(36)

### Trig values of 54 degrees

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