<s>[INST] Prove the following trigonometric identity without a calculator involved 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 $0$.
 [/INST] We may use formulas, mainly double angle formula and Chebyshev Polynomials, to simplify this problem:
$$\cos(2x)=2\cos^2(x)-1$$
$$\cos(4x)=2\cos^2(2x)-1=8\cos^4(x)-8\cos^2(x)+1$$
$$\cos(6x)=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1$$
$$\cos(8x)=64\cos^8(x)-128\cos^6(x)+22\cos^4(x)-16\cos^2(x)+1$$
Use $x=\frac{\pi}5$.
$$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}$$
$$=1+2\cos^2(x)-1+8\cos^4(x)-8\cos^2(x)+1+32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1+64\cos^8(x)-128\cos^6(x)+22\cos^4(x)-16\cos^2(x)+1$$
$$=1-4\cos^2(x)-18\cos^4(x)-96\cos^6(x)+64\cos^8(x)$$
We know that $\cos(\frac{\pi}5)=\frac{1+\sqrt{5}}4$.  Plug this in:
$$=1-4\left(\frac{1+\sqrt{5}}4\right)^2-18\left(\frac{1+\sqrt{5}}4\right)^4-96\left(\frac{1+\sqrt{5}}4\right)^6+64\left(\frac{1+\sqrt{5}}4\right)^8$$
Avoiding the messy stuff, I'll leave it to the reader to determine if the above is equal to zero.
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