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

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.