How can one prove:cos⁡(π7)+cos⁡(3π7)+cos⁡(5π7)=12

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ImpudgeIntemnect · Accepted Answer

Hint: start with eiπ7=cos⁡(π7)+isin⁡(π7) and the fact that the lhs is a 7th root of -1.Let u=eiπ7, then we want to find R(u+u3+u5)Then we have u7=−1 so u6−u5+u4−u3+u2−u+1=0Re-arranging this we get:u6+u4+u2+1=u5+u3+uIf a=u+u3+u5 then this becomes ua+1=a, and rearranging this gives a(1−u)=1, or a=11−uSo all we have to do is find R(11−u)11−u=11−cos⁡(π7)−isin⁡(π7)=1−cos⁡(π7)+isin⁡(π7)2−2cos⁡(π7)soR(11−u)=1−cos⁡(π7)2−2cos⁡(π7)=12

Rosalind Barker · Accepted Answer

To compute s=cos⁡(π7)+cos⁡(3π7)+cos⁡(5π7), one can continue the sum and considert=∑k=17cos⁡(2k−17π)Since cos⁡(2π−α)=cos⁡(α) for every α and cos⁡(π)=−1, t=2s−1. Now, t is the real part ofu=eiπ7+e3iπ7+e5iπ7+e7iπ7+e9iπ7+e11iπ7+e13iπ7and u=z+z3+z5+z7+z9+z11+z13 with z=eiπ7. This is a geometric sum with ratio z2≠1 hence u=z1−z141−z2.Finally, z14=e14iπ7=1 hence u=0. Thus, t=0, and s=12This is the n=3 case of the identity, valid for every n≥1∑k=1ncos⁡(2k−12n+1π)=12

How can one prove: \cos(\pi/7)+\cos(3\pi/7)+\cos(5\pi/7)=1/2

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