Proving cos⁡2π7+cos⁡4π7+cos⁡6π7=−12 Prove the identity 8cos4θ−4cos3θ−8cos2θ+3cos⁡θ+1=cos⁡4θ−cos⁡3θ If 7θ is a multiple of 2π, Show that cos⁡4θ=cos⁡3θ and deduce, cos⁡2π7+cos⁡4π7+cos⁡6π7=−12

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Proving cos⁡2π7+cos⁡4π7+cos⁡6π7=−12  Prove the identity  8cos4θ−4cos3θ−8cos2θ+3cos⁡θ+1=cos⁡4θ−cos⁡3θ  If 7θ is a multiple of 2π, Show that cos⁡4θ=cos⁡3θ and deduce,  cos⁡2π7+cos⁡4π7+cos⁡6π7=−12

Amari Larsen · Accepted Answer

cos⁡0π7,cos⁡2π7,cos⁡4π7,cos⁡6π7 are distinct roots of the fourth order polynomial  P(x)=8x4−4x3−8x2+3x+1  So P(x) can be re-written  P(x)=8(x−cos⁡0π7)(x−cos⁡2π7)(x−cos⁡4π7)(x−cos⁡6π7)  Therefore looking at x3 coefficient gives  cos⁡0π7+cos⁡2π7+cos⁡4π7+cos⁡6π7=48=12   Actually, for any n≥2,  Pn(x)=Tn(x)−Tn−1(x)=2n−1xn−2n−2xn−1+…   Where Tn is the nth Chebyshev polynomial.  So  Pn(cos⁡(x))=cos⁡(nx)−cos⁡((n−1)x)   And Likewise,  ∑k=0n−1cos⁡2kπ2n−1=12

Proving \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2} Prove the identity 8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta If

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