Monday Math 139

Prove that for positive integers k, m, and n, that
$\sin\frac{k\pi}n\sin\frac{m\pi}n=\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n$.

First, we distribute the $\sin\frac{\pi}n$ into the sum:
$\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n=\sum_{i=1}^{m}\left(\sin\frac{(k-m+2i-1)\pi}n\sin\frac{\pi}n\right)$;
next, we use the product identity for sines:
$\sin\alpha\sin\beta=\frac12\left(\cos(\alpha-\beta)-\cos(\alpha+\beta)\right)$, which means for the terms in the sum, we have
$\begin{array}{rcl}\sum_{i=1}^{m}\left(\sin\frac{(k-m+2i-1)\pi}n\sin\frac{\pi}n\right)&=&\sum_{i=1}^{m}\frac12\left[\cos\left(\frac{(k-m+2i-1)\pi}n-\frac{\pi}n\right)-\cos\left(\frac{(k-m+2i-1)\pi}n+\frac{\pi}n\right)\right]\\&=&\frac12\sum_{i=1}^{m}\left[\cos\left(\frac{(k-m+2i-2)\pi}n\right)-\cos\left(\frac{(k-m+2i)\pi}n\right)\right]\end{array}$.
Now, notice that this sum is
$\left[\cos\left(\frac{(k-m)\pi}n\right)-\cos\left(\frac{(k-m+2)\pi}n\right)\right]+\left[\cos\left(\frac{(k-m)\pi+2}n\right)-\cos\left(\frac{(k-m)\pi+4}n\right)\right]\\\;\;+\cdots+\left[\cos\left(\frac{(k-m+2m-2)\pi}n\right)-\cos\left(\frac{(k-m+2m)\pi}n\right)\right]$, and thus is a telescoping series:
$\begin{array}{rcl}\sum_{i=1}^{m}\left(\sin\frac{(k-m+2i-1)\pi}n\sin\frac{\pi}n\right)&=&\frac12\sum_{i=1}^{m}\left[\cos\left(\frac{(k-m+2i-2)\pi}n\right)-\cos\left(\frac{(k-m+2i)\pi}n\right)\right]\\&=&\frac12\left[\cos\left(\frac{(k-m)\pi}n\right)-\cos\left(\frac{(k-m+2m)\pi}n\right)\right]\\&=&\frac12\left[\cos\left(\frac{(k-m)\pi}n\right)-\cos\left(\frac{(k+m)\pi}n\right)\right]\\\sum_{i=1}^{m}\left(\sin\frac{(k-m+2i-1)\pi}n\sin\frac{\pi}n\right)&=&\frac12\left[\cos\left(\frac{k\pi}n-\frac{m\pi}n\right)-\cos\left(\frac{k\pi}n+\frac{m\pi}n\right)\right]\end{array}$,
and we can use our product rule for sines again, in the reverse direction; with $\alpha=\frac{k\pi}n$, and $\beta=\frac{m\pi}n$, we see
$\begin{array}{rcl}\sum_{i=1}^{m}\left(\sin\frac{(k-m+2i-1)\pi}n\sin\frac{\pi}n\right)&=&\frac12\left[\cos\left(\frac{k\pi}n-\frac{m\pi}n\right)-\cos\left(\frac{k\pi}n+\frac{m\pi}n\right)\right]\\\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n&=&\sin\frac{k\pi}n\sin\frac{m\pi}n\end{array}$,
and we have our proof.

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One Response to “Monday Math 139”

1. Monday Math 141 « Twisted One 151's Weblog Says:

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