1. ## series

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

For the first part , i got

$f(r+1)-f(r)=\cos (2r+2)\theta-\cos 2r\theta$ .. is this correct .

But from here , i couldn't relate to the second problem .

2. Originally Posted by thereddevils
If $f(r)=\cos 2r\theta$ , simplify $f(r+1)-f(r)$ . Use your result to find the sum of the first n terms of the series : $\sin {\color{red}3} \theta+\sin 5\theta+\sin 7\theta+...$ Mr F says: Correction in red.

For the first part , i got

$f(r+1)-f(r)=\cos (2r+2)\theta-\cos 2r\theta$ .. is this correct .

But from here , i couldn't relate to the second problem .
When the first part is answered completely the second part should become clearer:

$f(r + 1) = \cos(2[r+1] \theta) = \cos(2r \theta + 2 \theta) = \cos (2 r \theta) \cos (2 \theta) - \sin (2 r \theta) \sin (2 \theta)$.

Therefore:

$f(r + 1) - f(r) = \cos (2 r \theta) \cos (2 \theta) - \sin (2 r \theta) \sin (2 \theta) - \cos (2 r \theta)$

$= \cos (2 r \theta) [\cos (2 \theta) - 1] - \sin (2 r \theta) \sin (2 \theta)$

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

$= - 2 \sin (\theta) [ \cos (2 r \theta) \sin (\theta ) + \sin (2 r \theta) \cos (\theta)] = -2 \sin (\theta) \sin (2 r \theta + \theta)$

$= -2 \sin (\theta) \sin ([2 r + 1]\theta)$.

By the way, the result in the second part of the question has a typo which I have corrected.