## Statement

The exact value of the trigonometry functions of $ 45 \degree $ are:

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

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

$$ \tan(45 \degree) = 1 $$

## Proof

Construct triangle $ ABC $ with $ \angle A = \angle C = 45 \degree $ and $ \angle B = 90 \degree $, like the image below.

Note that this triangle is an isosceles triangle with $ AB = BC $. Now call those sides $ m $.

Calculate $ AC $ using the Pythagorean theorem:

$$ AC = \sqrt{AB^2 + BC^2} = \sqrt{m^2 + m^2} = \sqrt{2m^2} = m\sqrt{2} $$

### Sine

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

$$ \sin(45 \degree) = \frac{m}{m\sqrt{2}} = \frac{1}{\sqrt{2}} = \tfrac{1}{2}\sqrt{2} $$

### Cosine

From the triangle, note that $ \cos(\angle A) = \dfrac{AB}{AC} $, so:

$$ \cos(45 \degree) = \frac{m}{m\sqrt{2}} = \frac{1}{\sqrt{2}} = \tfrac{1}{2}\sqrt{2} $$

### Tangent

From the triangle, note that $ \tan(\angle A) = \dfrac{BC}{AB} $, so:

$$ \tan(45 \degree) = \frac{m}{m} = 1 $$