TITLE: Evaluate $\cos 18^\circ$ without using the calculator
QUESTION [13 upvotes]: 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
$$\cos 18^\circ=\frac{1}{4}\sqrt{10+2\sqrt{5}}$$

REPLY [15 votes]: Consider the isosceles triangle pictured below: 

By considering similar triangles, one has ${x\over 1}={1\over x-1}$.  From this, it follows that $x={1+\sqrt 5\over 2}$. Drop a perpendicular from the top vertex to the base below. One sees that $\sin(18^\circ)={1\over 1+\sqrt 5}={\sqrt 5-1\over 4}$. Now use the Pythagorean Identity to find $\cos(18^\circ)$.