Statement
The exact value of the trigonometry functions of $ 54 \degree $ are:
$$ \sin(54 \degree) = \tfrac14 \sqrt{6 + 2\sqrt5} $$
$$ \cos(54 \degree) = \tfrac14 \sqrt{10 - 2\sqrt5} $$
$$ \tan(54 \degree) = \tfrac15 \sqrt{25 + 10\sqrt5} $$
Sine
Use the sum of angles formula for sine:
$$ \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) $$
When substituting $ 18 \degree $ for $ \alpha $ and $ 36 \degree $ for $ \beta $, you get:
$$ \sin(54 \degree) = \sin(18 \degree + 36 \degree) = \sin(18 \degree) \cos(36 \degree) + \cos(18 \degree) \sin(36 \degree) $$
Recall the exact trig values of $ 18 \degree $ and the exact trig values of $ 36 \degree $:
$$ \sin(18 \degree) = \tfrac14 \sqrt{6 - 2\sqrt5} $$
$$ \cos(18 \degree) = \tfrac14 \sqrt{10 + 2\sqrt5} $$
$$ \sin(36 \degree) = \tfrac14 \sqrt{10 - 2\sqrt5} $$
$$ \cos(36 \degree) = \tfrac14 \sqrt{6 + 2\sqrt5} $$
Now substitute these values, and simplify:
$$\begin{aligned}\sin(54 \degree) &= \cos(36 \degree) \sin(18 \degree) + \sin(36 \degree) \cos(18 \degree) \newline&= \tfrac14 \sqrt{6 + 2\sqrt5} \cdot \tfrac14 \sqrt{6 - 2\sqrt5} + \tfrac14 \sqrt{10 - 2\sqrt5} \cdot \tfrac14 \sqrt{10 + 2\sqrt5} \newline&= \tfrac{1}{16} \sqrt{\left(6 + 2\sqrt5\right)\left(6 - 2\sqrt5\right)} + \tfrac{1}{16} \sqrt{\left(10 - 2\sqrt5\right)\left(10 + 2\sqrt5\right)} \newline&= \tfrac{1}{16} \sqrt{36 - 20} + \tfrac{1}{16} \sqrt{100 - 20} \newline&= \tfrac{1}{16} \sqrt{16} + \tfrac{1}{16} \sqrt{80} \newline&= \tfrac14 + \tfrac14 \sqrt5 \newline&= \tfrac14 \left(\sqrt5 + 1\right) \newline&= \tfrac14 \sqrt{\left(\sqrt5 + 1\right)^2} \newline&= \tfrac14 \sqrt{6 + 2\sqrt5}\end{aligned}$$
Cosine
Use the sum of angles formula for cosine:
$$ \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) $$
When substituting $ 36 \degree $ for $ \alpha $ and $ 18 \degree $ for $ \beta $, you get:
$$ \cos(54 \degree) = \cos(36 \degree + 18 \degree) = \cos(36 \degree) \cos(18 \degree) - \sin(36 \degree) \sin(18 \degree) $$
Recall the exact trig values of $ 18 \degree $ and the exact trig values of $ 36 \degree $:
$$ \sin(18 \degree) = \tfrac{1}{4} \sqrt{6 - 2\sqrt5} $$
$$ \cos(18 \degree) = \tfrac{1}{4} \sqrt{10 + 2\sqrt5} $$
$$ \sin(36 \degree) = \tfrac14 \sqrt{10 - 2\sqrt5} $$
$$ \cos(36 \degree) = \tfrac14 \sqrt{6 + 2\sqrt5} $$
Now substitute these values, and simplify:
$$\begin{aligned}\cos(54 \degree) &= \cos(36 \degree) \cos(18 \degree) - \sin(36 \degree) \sin(18 \degree) \newline&= \tfrac14 \sqrt{6 + 2\sqrt5} \cdot \tfrac14 \sqrt{10 + 2\sqrt5} - \tfrac14 \sqrt{10 - 2\sqrt5} \cdot \tfrac14 \sqrt{6 - 2\sqrt5} \newline&= \tfrac{1}{16} \sqrt{\left(6 + 2\sqrt5\right)\left(10 + 2\sqrt5\right)} - \tfrac{1}{16} \sqrt{\left(10 - 2\sqrt5\right)\left(6 - 2\sqrt5\right)} \newline&= \tfrac{1}{16} \sqrt{60 + 12\sqrt5 + 20\sqrt5 + 20} - \tfrac{1}{16} \sqrt{60 - 20\sqrt5 - 12\sqrt5 + 20} \newline&= \tfrac{1}{16} \sqrt{80 + 32\sqrt5} - \tfrac{1}{16} \sqrt{80 - 32\sqrt5} \newline&= \tfrac14 \sqrt{5 + 2\sqrt5} - \tfrac14 \sqrt{5 - 2\sqrt5} \newline&= \tfrac14 \left(\sqrt{5 + 2\sqrt5} - \sqrt{5 - 2\sqrt5}\right) \newline&= \tfrac14 \sqrt{\left(\sqrt{5 + 2\sqrt5} - \sqrt{5 - 2\sqrt5}\right)^2} \newline&= \tfrac14 \sqrt{5 + 2\sqrt5 - 2\sqrt{\left(5 + 2\sqrt5\right)\left(5 - 2\sqrt5\right)} + 5 - 2\sqrt5} \newline&= \tfrac14 \sqrt{10 - 2\sqrt{25 - 20}} \newline&= \tfrac14 \sqrt{10 - 2\sqrt5}\end{aligned}$$
Tangent
Use that tangent is sine divided by cosine and simplify.
$$\begin{aligned}\tan(54 \degree) &= \frac{\sin(54 \degree)}{\cos(54 \degree)} \newline&= \frac{\tfrac14 \sqrt{6 + 2\sqrt5}}{\tfrac14 \sqrt{10 - 2\sqrt5}} \newline&= \sqrt{\frac{6 + 2\sqrt5}{10 - 2\sqrt5}} \newline&= \sqrt{\frac{6 + 2\sqrt5}{10 - 2\sqrt5} \cdot \frac{10 + 2\sqrt5}{10 + 2\sqrt5}} \newline&= \sqrt{\frac{60 + 12\sqrt5 + 20\sqrt5 + 20}{100 - 20}} \newline&= \sqrt{\frac{80 + 32\sqrt5}{80}} \newline&= \sqrt{\frac{5 + 2\sqrt5}{5}} \newline&= \frac{\sqrt{5 + 2\sqrt5}}{\sqrt5} \newline&= \tfrac15 \sqrt5 \cdot \sqrt{5 + 2\sqrt5} \newline&= \tfrac15 \sqrt{25 + 10\sqrt5}\end{aligned}$$