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Summation: trigonometric identity

hxthanh

New member
Sep 20, 2013
16
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 $x$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Re: Summation trigonometry identity

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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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.