shen07
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$$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$
The discussion centers on the mathematical proof for the sum of cosines up to n terms, specifically represented by the equation $$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$. Participants utilized the substitution $$z=exp(ix)$$ to derive the result, leading to the expression $$(1+z)^n = 2^n cos^n (x/2) cos (nx/2)$$. The final answer is confirmed as the real part of the derived expression, showcasing a clear method for solving the problem.
PREREQUISITESMathematicians, physics students, and anyone interested in advanced trigonometric identities and their proofs will benefit from this discussion.
shen07 said:Got the answer..