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