Statement
The exact value of the trigonometry functions of $ 3 \degree $ are:
$$ \sin(3 \degree) = \tfrac14 \sqrt{8 - \sqrt3 - \sqrt{15} - \sqrt{10 - 2\sqrt5}} $$
$$ \cos(3 \degree) = \tfrac14 \sqrt{8 + \sqrt3 + \sqrt{15} + \sqrt{10 - 2\sqrt5}} $$
$$ \tan(3 \degree) = \sqrt{\frac{8 - \sqrt3 - \sqrt{15} - \sqrt{10 - 2\sqrt5}}{8 + \sqrt3 + \sqrt{15} + \sqrt{10 - 2\sqrt5}}} $$
Sine
Use the difference of angles formula for cosine:
$$ \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \sin(\beta) \cos(\alpha) $$
When substituting $ 18 \degree $ for $ \alpha $ and $ 15 \degree $ for $ \beta $, you get:
$$ \sin(3 \degree) = \sin(18 \degree - 15 \degree) = \sin(18 \degree) \cos(15 \degree) - \sin(15 \degree) \cos(18 \degree) $$
Recall the exact trig values of $ 18 \degree $ and the exact trig values of $ 15 \degree $:
$$ \sin(18 \degree) = \tfrac14 \sqrt{6 - 2\sqrt5} $$
$$ \cos(18 \degree) = \tfrac14 \sqrt{10 + 2\sqrt5} $$
$$ \sin(15 \degree) = \tfrac14 \sqrt{8 - 4\sqrt3} $$
$$ \cos(15 \degree) = \tfrac14 \sqrt{8 + 4\sqrt3} $$
Now substitute these values, and simplify:
$$\begin{aligned}\sin(3 \degree) &= \sin(18 \degree) \cos(15 \degree) - \cos(18 \degree) \sin(15 \degree) \newline&= \tfrac14 \sqrt{6 - 2\sqrt5} \cdot \tfrac14 \sqrt{8 + 4\sqrt3} - \tfrac14 \sqrt{10 + 2\sqrt5} \cdot \tfrac14 \sqrt{8 - 4\sqrt3} \newline&= \tfrac{1}{16} \sqrt{\left(6 - 2\sqrt5\right)\left(8 + 4\sqrt3\right)} - \tfrac{1}{16} \sqrt{\left(10 + 2\sqrt5\right)\left(8 - 4\sqrt3\right)} \newline&= \tfrac{1}{16} \sqrt{48 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15}} - \tfrac{1}{16} \sqrt{80 - 40\sqrt3 + 16\sqrt5 - 8\sqrt{15}} \newline&= \tfrac18 \sqrt{12 + 6\sqrt3 - 4\sqrt5 - 2\sqrt{15}} - \tfrac18 \sqrt{20 - 10\sqrt3 + 4\sqrt5 - 2\sqrt{15}} \newline&= \tfrac18 \left(\sqrt{12 + 6\sqrt3 - 4\sqrt5 - 2\sqrt{15}} - \sqrt{20 - 10\sqrt3 + 4\sqrt5 - 2\sqrt{15}}\right) \newline&= \tfrac18 \sqrt{\left(\sqrt{12 + 6\sqrt3 - 4\sqrt5 - 2\sqrt{15}} - \sqrt{20 - 10\sqrt3 + 4\sqrt5 - 2\sqrt{15}}\right)^2} \newline&= \tfrac18 \sqrt{12 + 6\sqrt3 - 4\sqrt5 - 2\sqrt{15} + 20 - 10\sqrt3 + 4\sqrt5 - 2\sqrt{15} - 2\sqrt{\left(12 + 6\sqrt3 - 4\sqrt5 - 2\sqrt{15}\right)\left(20 - 10\sqrt3 + 4\sqrt5 - 2\sqrt{15}\right)}} \newline&= \tfrac18 \sqrt{32 - 4\sqrt3 - 4\sqrt{15} - 2\sqrt{240 - 120\sqrt3 + 48\sqrt5 - 24\sqrt{15} + 120\sqrt3 - 180 + 24\sqrt{15} - 36\sqrt5 - 80\sqrt5 + 40\sqrt{15} - 80 + 40\sqrt3 - 40\sqrt{15} + 60\sqrt5 - 40\sqrt3 + 60}} \newline&= \tfrac18 \sqrt{32 - 4\sqrt3 - 4\sqrt{15} - 2\sqrt{40 - 8\sqrt5}} \newline&= \tfrac18 \sqrt{32 - 4\sqrt3 - 4\sqrt{15} - 4\sqrt{10 - 2\sqrt5}} \newline&= \tfrac14 \sqrt{8 - \sqrt3 - \sqrt{15} - \sqrt{10 - 2\sqrt5}}\end{aligned}$$
Cosine
Use the difference of angles formula for cosine:
$$ \cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta) $$
When substituting $ 18 \degree $ for $ \alpha $ and $ 15 \degree $ for $ \beta $, you get:
$$ \cos(3 \degree) = \cos(18 \degree - 15 \degree) = \cos(18 \degree) \cos(15 \degree) + \sin(18 \degree) \sin(15 \degree) $$
Recall the exact trig values of $ 18 \degree $ and the exact trig values of $ 15 \degree $:
$$ \sin(18 \degree) = \tfrac14 \sqrt{6 - 2\sqrt5} $$
$$ \cos(18 \degree) = \tfrac14 \sqrt{10 + 2\sqrt5} $$
$$ \sin(15 \degree) = \tfrac14 \sqrt{8 - 4\sqrt3} $$
$$ \cos(15 \degree) = \tfrac14 \sqrt{8 + 4\sqrt3} $$
Now substitute these values, and simplify:
$$\begin{aligned}\cos(3 \degree) &= \cos(18 \degree) \cos(15 \degree) + \sin(18 \degree) \sin(15 \degree) \newline&= \tfrac14 \sqrt{10 + 2\sqrt5} \cdot \tfrac14 \sqrt{8 + 4\sqrt3} + \tfrac14 \sqrt{6 - 2\sqrt5} \cdot \tfrac14 \sqrt{8 - 4\sqrt3} \newline&= \tfrac{1}{16} \sqrt{\left(10 + 2\sqrt5\right)\left(8 + 4\sqrt3\right)} + \tfrac{1}{16} \sqrt{\left(6 - 2\sqrt5\right)\left(8 - 4\sqrt3\right)} \newline&= \tfrac{1}{16} \sqrt{80 + 40\sqrt3 + 16\sqrt5 + 8\sqrt{15}} + \tfrac{1}{16} \sqrt{48 - 24\sqrt3 - 16\sqrt5 + 8\sqrt{15}} \newline&= \tfrac{1}{8} \sqrt{20 + 10\sqrt3 + 4\sqrt5 + 2\sqrt{15}} + \tfrac{1}{8} \sqrt{12 - 6\sqrt3 - 4\sqrt5 + 2\sqrt{15}} \newline&= \tfrac{1}{8} \left(\sqrt{20 + 10\sqrt3 + 4\sqrt5 + 2\sqrt{15}} + \sqrt{12 - 6\sqrt3 - 4\sqrt5 + 2\sqrt{15}}\right) \newline&= \tfrac{1}{8} \sqrt{\left(\sqrt{20 + 10\sqrt3 + 4\sqrt5 + 2\sqrt{15}} + \sqrt{12 - 6\sqrt3 - 4\sqrt5 + 2\sqrt{15}}\right)^2} \newline&= \tfrac{1}{8} \sqrt{20 + 10\sqrt3 + 4\sqrt5 + 2\sqrt{15} + 12 - 6\sqrt3 - 4\sqrt5 + 2\sqrt{15} + 2\sqrt{\left(20 + 10\sqrt3 + 4\sqrt5 + 2\sqrt{15}\right)\left(12 - 6\sqrt3 - 4\sqrt5 + 2\sqrt{15}\right)}} \newline&= \tfrac{1}{8} \sqrt{32 + 4\sqrt3 + 4\sqrt{15} + 2\sqrt{240 - 120\sqrt3 - 80\sqrt5 + 40\sqrt{15} + 120\sqrt3 - 180 - 40\sqrt{15} + 60\sqrt5 + 48\sqrt5 - 24\sqrt{15} - 80 + 40\sqrt3 + 24\sqrt{15} - 36\sqrt5 - 40\sqrt3 + 60}} \newline&= \tfrac{1}{8} \sqrt{32 + 4\sqrt3 + 4\sqrt{15} + 2\sqrt{40 - 8\sqrt5}} \newline&= \tfrac{1}{8} \sqrt{32 + 4\sqrt3 + 4\sqrt{15} + 4\sqrt{10 - 2\sqrt5}} \newline&= \tfrac{1}{4} \sqrt{8 + \sqrt3 + \sqrt{15} + \sqrt{10 - 2\sqrt5}} \newline\end{aligned}$$
Tangent
Use that tangent is sine divided by cosine and simplify.
$$\begin{aligned}\tan(3 \degree) &= \frac{\sin(3 \degree)}{\cos(3 \degree)} \newline&= \frac{\tfrac14 \sqrt{8 - \sqrt3 - \sqrt{15} - \sqrt{10 - 2\sqrt5}}}{\tfrac14 \sqrt{8 + \sqrt3 + \sqrt{15} + \sqrt{10 - 2\sqrt5}}} \newline&= \sqrt{\frac{8 - \sqrt3 - \sqrt{15} - \sqrt{10 - 2\sqrt5}}{8 + \sqrt3 + \sqrt{15} + \sqrt{10 - 2\sqrt5}}}\end{aligned}$$