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Prove that (not use induction) $\displaystyle\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} = \frac{1+(-1)^n}{2\cos^nx} + \dfrac{2\sin\big(\lfloor\frac{n+1}{2}\rfloor x\big) \cos\big(\lfloor\frac{n+2}{2}\rfloor x\big)} {\sin x\cos^n x} \qquad\qquad (\frac{2x}{\pi}\not\in \mathbb Z)$

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  • $\begingroup$ Why should we not use induction? $\endgroup$ – GEdgar Sep 21 '13 at 13:10
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HINT:

Euler Formula says $e^{iy}=\cos y+\sin y,$

$\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} =$Re $\left(\sum_{k=0}^n \left(\frac{e^{ix}}{\cos x}\right)^k\right)$

Now, $\sum_{k=0}^n \left(\frac{e^{ix}}{\cos x}\right)^k$ is a Geometric Series

Now, deal the even & the odd cases of $n$ separately

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