MHB Summation: trigonometric identity

AI Thread Summary
The discussion centers around proving the trigonometric identity involving a summation of cosine terms divided by powers of cosine. Participants suggest using mathematical induction to approach the proof, recommending separate treatment for even and odd cases of n. The original poster clarifies that the question is intended as a challenge rather than a request for assistance. The conversation emphasizes the complexity of the identity and the need for careful consideration in the proof process. Overall, the thread highlights the intricacies of trigonometric identities and the methods used to prove them.
hxthanh
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Prove that:
$\displaystyle\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} = \frac{1+(-1)^n}{2\cos^nx} + \dfrac{2\sin\big(\lfloor\frac{n+1}{2}\rfloor x\big) \cos\big(\lfloor\frac{n+2}{2}\rfloor x\big)} {\sin x\cos^n x} \qquad\qquad (\frac{2x}{\pi}\not\in \mathbb Z)$

*note: $\lfloor x\rfloor$ is floor function of $x$
 
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Re: Summation trigonometry identity

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

*note: $\lfloor x\rfloor$ is floor function of $x$
Have you tried proving this by induction? Hint: you may find it easier to treat the cases $n$ even and $n$ odd separately.
 
Re: Summation trigonometry identity

I have learned from the OP that this question is meant as a challenge rather than seeking help, so it has been moved accordingly.
 
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