Trig values of 30 and 60 degrees

Statement

The exact value of the trigonometry functions of $30 \degree$ and $60 \degree$ are:

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

$$\cos(30 \degree) = \tfrac{1}{2}\sqrt{3}$$

$$\tan(30 \degree) = \tfrac{1}{3}\sqrt{3}$$

$$\sin(60 \degree) = \tfrac{1}{2}\sqrt{3}$$

$$\cos(60 \degree) = \tfrac{1}{2}$$

$$\tan(60 \degree) = \sqrt{3}$$

Proof

Construct triangle $ABC$ with $\angle A = \angle B = \angle C = 60 \degree$. Then draw altitude $CD$ on $AB$ so that $\angle C_1 = \angle C_2 = 30 \degree$, like the image below.

Let $BD = m$. Because this is an equilateral triangle, $BC = 2m$.

Now find $CD$ from the Pythagorean theorem:

$$CD = \sqrt{BC^2 - BD^2} = \sqrt{(2m)^2 - m^2} = \sqrt{4m^2 - m^2} = \sqrt{3m^2} = m\sqrt{3}$$

Sine 30

From the triangle, note that $\sin(\angle C_2) = \dfrac{BD}{BC}$, so:

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

Cosine 30

From the triangle, note that $\cos(\angle C_2) = \dfrac{CD}{BC}$, so:

$$\cos(30 \degree) = \frac{m\sqrt{3}}{2m} = \tfrac{1}{2}\sqrt{3}$$

Tangent 30

From the triangle, note that $\tan(\angle C_2) = \dfrac{BD}{CD}$, so:

$$\tan(30 \degree) = \frac{m}{m\sqrt{3}} = \frac{1}{\sqrt{3}} = \tfrac{1}{3}\sqrt{3}$$

Sine 60

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

$$\sin(60 \degree) = \frac{m\sqrt{3}}{2m} = \tfrac{1}{2}\sqrt{3}$$

Cosine 60

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

$$\cos(60 \degree) = \frac{m}{2m} = \tfrac{1}{2}$$

Tangent 60

From the triangle, note that $\tan(\angle B) = \dfrac{CD}{BD}$, so:

$$\tan(60 \degree) = \frac{m\sqrt{3}}{m} = \sqrt{3}$$