Statement
The exact value of the trigonometry functions of $ 15 \degree $ are:
$$ \sin(15 \degree) = \tfrac{1}{4} \sqrt{8 - 4\sqrt3} $$
$$ \cos(15 \degree) = \tfrac{1}{4} \sqrt{8 + 4\sqrt3} $$
$$ \tan(15 \degree) = \sqrt{7 - 4\sqrt3} $$
Sine
Use the difference of angles formula for sine:
$$ \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \sin(\beta) \cos(\alpha) $$
When substituting $ 45 \degree $ for $ \alpha $ and $ 30 \degree $ for $ \beta $, you get:
$$ \sin(15 \degree) = \sin(45 \degree - 30 \degree) = \sin(45 \degree) \cos(30 \degree) - \sin(30 \degree) \cos(45 \degree) $$
Recall the exact trig values of $ 30 \degree $ and the exact trig values of $ 45 \degree $:
$$ \sin(30 \degree) = \tfrac{1}{2} $$
$$ \cos(30 \degree) = \tfrac{1}{2} \sqrt3 $$
$$ \sin(45 \degree) = \tfrac{1}{2} \sqrt2 $$
$$ \cos(45 \degree) = \tfrac{1}{2} \sqrt2 $$
Now substitute these values, and simplify:
$$\begin{aligned}\sin(15 \degree) &= \sin(45 \degree) \cos(30 \degree) - \sin(30 \degree) \cos(45 \degree) \newline&= \tfrac12 \sqrt2 \cdot \tfrac12 \sqrt3 - \tfrac12 \cdot \tfrac12 \sqrt2 \newline&= \tfrac14 \sqrt6 - \tfrac14 \sqrt2 \newline&= \tfrac14 \sqrt{\left(\sqrt6 - \sqrt2\right)^2} \newline&= \tfrac14 \sqrt{6 - 2\sqrt{12} + 2} \newline&= \tfrac14 \sqrt{8 - 4\sqrt3}\end{aligned}$$
Cosine
Use the difference of angles formula for cosine:
$$ \cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta) $$
When substituting $ 45 \degree $ for $ \alpha $ and $ 30 \degree $ for $ \beta $, you get:
$$ \cos(15 \degree) = \cos(45 \degree - 30 \degree) = \cos(45 \degree) \cos(30 \degree) + \sin(45 \degree) \sin(30 \degree) $$
Recall the exact trig values of $ 30 \degree $ and the exact trig values of $ 45 \degree $:
$$ \sin(30 \degree) = \tfrac{1}{2} $$
$$ \cos(30 \degree) = \tfrac{1}{2} \sqrt3 $$
$$ \sin(45 \degree) = \tfrac{1}{2} \sqrt2 $$
$$ \cos(45 \degree) = \tfrac{1}{2} \sqrt2 $$
Now substitute these values, and simplify:
$$\begin{aligned}\cos(15 \degree) &= \cos(45 \degree) \cos(30 \degree) + \sin(45 \degree) \sin(30 \degree) \newline&= \tfrac12 \sqrt2 \cdot \tfrac12 \sqrt3 + \tfrac12 \sqrt2 \cdot \tfrac12 \newline&= \tfrac14 \sqrt6 + \tfrac14 \sqrt2 \newline&= \tfrac14 \sqrt{\left(\sqrt6 + \sqrt2\right)^2} \newline&= \tfrac14 \sqrt{6 + 2\sqrt{12} + 2} \newline&= \tfrac14 \sqrt{8 + 4\sqrt3}\end{aligned}$$
Tangent
Use that tangent is sine divided by cosine and simplify.
$$\begin{aligned}\tan(15 \degree) &= \frac{\sin(15 \degree)}{\cos(15 \degree)} \newline&= \frac{\tfrac14 \sqrt{8 - 4\sqrt3}}{\tfrac14 \sqrt{8 + 4\sqrt3}} \newline&= \sqrt{\frac{8 - 4\sqrt3}{8 + 4\sqrt3}} \newline&= \sqrt{\frac{8 - 4\sqrt3}{8 + 4\sqrt3} \cdot \frac{8 - 4\sqrt3}{8 - 4\sqrt3}} \newline&= \sqrt{\frac{64 - 64\sqrt3 + 48}{64 - 48}} \newline&= \sqrt{\frac{16(4 - 4\sqrt3 + 3)}{16}} \newline&= \sqrt{7 - 4\sqrt3} \newline\end{aligned}$$