Use the identity $\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$ to prove that:
$2\sin(j\theta)\sin(\frac{1}{2}\theta)=\cos((j-\frac{1}{2})(\theta))-\cos((j+\frac{1}{2})\theta).$
This seemed almost too easy, so I am wondering if I am missing something?
I just let $A = j\theta$ and $B = \frac{1}{2}\theta$, and substituted it directly into the identity. Since the question has asked me to use the identity, I am assuming that we can use it directly.
Have I missed anything?
The next part of the question asks:
Deduce that
$\sum^n_{j=1}\sin(j\theta) = \frac{\cos(\frac{1}{2}\theta)-\cos((n+\frac{1}{2})\theta)}{2\sin(\frac{1}{2}\theta)}, \text{ if $\theta$ is not a multiple of $2\pi$}$
Writing this out in general terms I see that the middle terms all cancel out (i have omitted the denominator here):
$\cos((j-\frac{1}{2})\theta)-\cos((j+\frac{1}{2})\theta) + \cos(((j+1)-\frac{1}{2})\theta)-\cos(((j+1)+\frac{1}{2})\theta) + \cos(((j+2)-\frac{1}{2})\theta)-\cos(((j+2)+\frac{1}{2})\theta)+\dotsb+ \cos((n-\frac{1}{2})\theta)-\cos((n+\frac{1}{2})\theta)$
I notice that if $\theta$ is a multiple of $2\pi$, then I would lose the 1st term, thus $\theta$ cannot be a multiple of $2\pi$.
Can anyone give me some direction to make this more mathematically rigorous? I feel like I am just a few steps away from answering this correctly.
Thanks in advance!