Evaluate $\cos 18^\circ$ without using the calculator I only know $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, $180^\circ$, $270^\circ$, and $360^\circ$ as standard angles but how can I prove that
$$\cos 18^\circ=\frac{1}{4}\sqrt{10+2\sqrt{5}}$$

Let $\theta = 18^\circ$. 
Then $2\theta = 36^\circ$ and $3\theta = 54^\circ$.
Note that $90^\circ-3\theta = 2\theta$.
Thus, $\sin(90^\circ - 3\theta) = \sin (2\theta)$ or $\cos(3\theta) = \sin(2\theta)$
So, $4\cos^3\theta-3\cos\theta = 2\sin\theta\cos\theta$.
Since $\cos \theta$ can't be zero, we can divide by it to obtain,
$4\cos^2\theta-3 = 2\sin\theta$
Now use $\cos^2 \theta = 1 - \sin^2 \theta$ to obtain a quadratic in $\sin \theta$.
Solve it and then disregarding the negative root (since $\sin 18^\circ$ can't be negative), solve for $\cos 18^\circ$.
