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:

    […] us combine the results from the previous two weeks (here and here). We found that for a regular n-gon with unit sides, the diagonals have lengths , […]

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