# Summation: trigonometric identity

#### hxthanh

##### New member
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
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.