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How to show that

if $0\le\theta\le2\pi$

$|\sum\limits_{n=1}^{p}{\sin{n\theta}}|\le\csc{\frac{\theta}{2}}$

for all integer p?

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1 Answer 1

up vote 2 down vote accepted

$$S_p = \sum_{n=1}^{p} \sin(n \theta) = \csc (\theta/2) \sin(p \theta/2) \sin((p+1) \theta/2)$$ To see why the above is true, multiply $\displaystyle \sum_{n=1}^{p} \sin(n \theta)$ by $\sin(\theta/2)$ and write it as a difference of cosines, and telescoping will give you the answer. Now, since $\sin(\alpha) \leq 1$, $\forall \alpha \in \mathbb{R}$ we hence get that $$S_p = \sum_{n=1}^{p} \sin(n \theta) \leq \csc (\theta/2)$$

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