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