1. ## Series expressions

By expressing $2\sin\theta\cos 2r\theta$ as the difference between two sines, show that
$\sin\theta \displaystyle \sum^n_{r=1}\cos 2r\theta=\sin n\theta\cos (n+1)\theta$
Evaluate $\displaystyle \sum^{100}_{r=1} \cos^2 \left(\frac{r \pi}{100}\right)$

My attempt:
$2\sin\theta\cos 2r\theta$
$=2\sin\theta(\cos^2 r\theta-\sin^2 r\theta)$
$=\sin (r+1)\theta\cos r\theta-2\sin\theta\sin^2 r\theta$
$=\frac{1}{2}\sin (2r+1)\theta -2\sin\theta(1-\cos^2 r\theta)$
$=\frac{1}{2}\sin (2r+1)\theta+\sin (r+1)\theta\cos r\theta -2\sin\theta$
$=\frac{1}{2}\sin (2r+1)\theta+\frac{1}{2}\sin (2r+1)\theta-2\sin\theta$
$=\sin (2r+1)\theta-2\sin\theta$

$\sin\theta \displaystyle \sum^n_{r=1}\cos 2r\theta=\sin n\theta\cos (n+1)\theta=\frac{1}{2} \displaystyle \sum^n_{r=1}\left(\sin (2r+1)\theta-2\sin\theta\right)$
$=\frac{1}{2}\left(\displaystyle \sum^n_{r=1} \sin (2r+1)\theta-2n\sin\theta\right)$
This is as far as I got.
Thanks!

2. I would start by

$2\sin{\theta}\cos{2r\theta} = 2\sin{\theta}(1 - 2\sin^2{r\theta})$

$= 2\sin{\theta} - 4\sin{\theta}\sin^2{r\theta}$.

3. Thanks, that is a much shorter way, but I'm still not sure how to continue other than
$\sin\theta \displaystyle \sum^n_{r=1} \cos 2r\theta=\sum^n_{r=1} \left(\sin\theta-2\sin\theta\sin^2r\theta\right)$
then
$\sin\theta\left(n - \displaystyle \sum^n_{r=1}\sin^2r\theta\right)$