Prove sum of $\cos(\pi/11)+\cos(3\pi/11)+...+\cos(9\pi/11)=1/2$ using Euler's formula
Solution 1:
Put $$S = \cos(π/11)+\cos(3π/11)+\cos(5π/11)+\cos(7π/11)+\cos(9π/11)$$
Then $$S = \cos(-π/11)+\cos(-3π/11)+\cos(-5π/11)+\cos(-7π/11)+\cos(-9π/11)$$ (because cos is even)
and of course $$ -1 = \cos (-11\pi/11)$$
Sum these all up and you get $$2S-1$$ as the sum of the real parts of the eleven eleventh-roots of unity, which is 0, and therefore $$S = 1/2$$