TITLE: Prove the following trigonometric identity without a calculator involved
QUESTION [25 upvotes]: I have to prove the following statement.

$$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}=0$$

I have tried to use the sum of angles formula for cosine, but didn't get to a point where I'd be able to show that it is equal to $0$.

REPLY [4 votes]: The given terms are projections of unit radii of a regular pentagon on x-axis. A sum of cyclic vectors or static equilibrium of vectors/forces acting on a point, it sums to zero. 
BTW and likewise, the y-axis projection sum
$$ 1+\sin{2\pi\over5}+\sin{4\pi\over5}+\sin{6\pi\over5}+\sin{8\pi\over5}  $$
also goes to zero.
Also we have the formula for sum of cosines of $n$ angles in arithmetic progression with common difference $ \beta$
$$  \dfrac{\sin n \beta/2 }{ \sin \beta/2} \cdot \cos \dfrac{\alpha_1 +\alpha_2}{2} $$
which also vanishes.