How to prove $2\cos\frac{8\pi}{11}+2\cos\frac{6\pi}{11}+ 2\cos\frac{4\pi}{11}+2\cos\frac{2\pi}{11}+1=2\cos\frac{\pi}{11}$

Question

I verified that the following identities hold in Maple:

$$2\cos\frac{8\pi}{11}+2\cos\frac{6\pi}{11}+ 2\cos\frac{4\pi}{11}+2\cos\frac{2\pi}{11}+1=2\cos\frac{\pi}{11} \tag{1}$$ $$\exp\frac{2\pi i}{11} - \exp\frac{9\pi i}{11}=2\cos\frac{2\pi}{11} \tag{2}$$

I tried to prove them but was not successful. How can we prove these identities? Thank you very much.

$\cos(\pi-x)=-\cos x$ for the right hand side and use http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression — lab bhattacharjee, Mar 22 '17 at 16:42

score 1 · Answer 1 · answered Mar 22 '17 at 15:57

$2\cos(8π/11)+2\cos(6π/11)+2\cos(4π/11)+2\cos(2π/11)+1$

$\cos(8π/11)+\cos(6π/11)+\cos(4π/11)+\cos(2π/11)+\cos 0+\cos(-2π/11)+\cos(-4π/11)+\cos(-6π/11)+\cos(-8π/11) $

$z = e^{\frac {2\pi}{11}i}$

$Re[z^4 + z^3 + z^2 + z + z^0 + z^{-1} + z^{-2} + z^{-3} + z^{-4}]$

$z^5+z^4 + z^3 + z^2 + z + z^0 + z^{-1} + z^{-2} + z^{-3} + z^{-4}+z^{-5} - (z^5 + z^{-5})$

$z^5+z^4 + z^3 + z^2 + z + z^0 + z^{-1} + z^{-2} + z^{-3} + z^{-4}+z^{-5} = \frac{z^{11}-1}{z^5(z-1)}\\ z^{11} =e^{2\pi i} = 1$

$z^5+z^4 + z^3 + z^2 + z + z^0 + z^{-1} + z^{-2} + z^{-3} + z^{-4}+z^{-5} = 0$

$- (z^5 + z^{-5}) = -2\cos \frac {10 \pi}{11}=2\cos\frac {\pi}{11}$

How to prove $2\cos\frac{8\pi}{11}+2\cos\frac{6\pi}{11}+ 2\cos\frac{4\pi}{11}+2\cos\frac{2\pi}{11}+1=2\cos\frac{\pi}{11}$

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