# Thread: Sum of series by method of differences

1. ## Sum of series by method of differences

If $f(r)\equiv\cos 2r\theta$, find $f(r)-f(r+1)$. Use yur result to find the sum of the first n terms of the series
$\sin 3\theta+\sin 5\theta+\sin 7\theta+...$

I've got $f(r)-f(r+1)\equiv\cos 2r\theta-\cos 2(r+1)\theta$
$\equiv\cos 2r\theta-\cos (2r\theta+2\theta)$
$\equiv\cos 2r\theta-\cos 2r\theta\cos 2\theta+\sin 2r\theta\sin 2\theta$
$\equiv\cos 2r\theta(1-\cos 2\theta)+\sin 2r\theta\sin 2\theta$
$\equiv2\cos 2r\theta\sin^2\theta+sin 2r\theta\sin 2\theta$
$\equiv\sin\theta(\cos 2r\theta\sin\theta+\sin 2r\theta\cos\theta)+\cos 2r\theta\sin\theta$
$\equiv\sin\theta(\sin(2r+1)\theta)+\cos 2r\theta\sin\theta$
Now I don't know how to continue, or how this relates to the series.
Thanks for any help!

2. Your calculations are almost all good, except that you make a mistake in the last 2 lines. It is actually

$\displaystylef(r)-f(r+1)\equiv\cos 2r\theta-\cos 2(r+1)\theta$
$\displaystyle\equiv\cos 2r\theta-\cos (2r\theta+2\theta)$
$\displaystyle\equiv\cos 2r\theta-\cos 2r\theta\cos 2\theta+\sin 2r\theta\sin 2\theta$
$\displaystyle\equiv\cos 2r\theta(1-\cos 2\theta)+\sin 2r\theta\sin 2\theta$
$\displaystyle\equiv2\cos 2r\theta\sin^2\theta+sin 2r\theta\sin 2\theta$
$\displaystyle\equiv2\cos 2r\theta\sin^2\theta+2 \sin 2r\theta\sin \theta\cos\theta$
$\displaystyle\equiv 2\sin\theta(\cos 2r\theta\sin\theta+ \sin 2r\theta\cos\theta)$
$\displaystyle\equiv 2\sin\theta(\sin((2r+1)\theta))$

This helps by getting that

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So now, the sum on top is

$$\displaystyle \sin 3\theta+\sin 5\theta+\sin 7\theta+... = \frac{1}{2\sin\theta}\left(\sum_{r = 1}^n (f(r)-f(r+1)\right)$$

Which is a telescoping sum which sums to

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