<s>[INST] Sums $\sum_{k=1}^n \sin(2k-1)\theta$, $\sum_{k=1}^n \sin^2(2k-1)\theta $ To prove: 
$1.$ $$\sum_{k=1}^n \sin(2k-1)\theta = \frac{\sin^2 n\theta}{\sin \theta}.$$
$2.$ $$\sum_{k=1}^n \sin^2(2k-1)\theta = \frac{n}{2} - \frac{\sin 4n\theta}{4\sin 2\theta}.$$
 [/INST] Alternatively, from $\cos(a+b)=\cos a\cos b-\sin a\sin b$ we get $$\sin a\sin b=\frac{\cos(b-a)-\cos(b+a)}2$$
Giving appropriate values to $a,b$, obtain a telescopical sum on the left. 
Let $a=x,b=(2k-1)x$. Then $$\eqalign{
  & \sin x\sin \left( {2k - 1} \right)x = \frac{{\cos (2k - 2)x - \cos 2kx}}{2}  \cr 
  & \sin x\sum\limits_{k = 1}^n {\sin \left( {2k - 1} \right)x}  = \frac{{1 - \cos 2nx}}{2} \cr} $$
so $$\sum\limits_{k = 1}^n {\sin \left( {2k - 1} \right)x}  = \frac{{{{\sin }^2}nx}}{{\sin x}}$$
For the second one, note that $${\sin ^2}(2k - 1)\theta  = \frac{{1 - \cos \left( {2k - 1} \right)2\theta }}{2}$$ so it suffices to find $$\sum\limits_{k = 1}^n {\cos \left( {2k - 1} \right)x} $$
And again, we can use $$\frac{{\sin \left( {a + b} \right) - \sin \left( {a - b} \right)}}{2} = \sin b\cos a$$ to get $$\sin x\cos \left( {2k - 1} \right)x = \frac{{\sin 2kx - \sin \left( {2k - 2} \right)x}}{2}$$ $$\sum\limits_{k = 1}^n {\cos \left( {2k - 1} \right)x}  = \frac{{\sin 2nx}}{{2\sin x}}$$
so $$\sum\limits_{k = 1}^n {{{\sin }^2}(2k - 1)x}  = \frac{n}{2} - \frac{{\sin 4nx}}{{4\sin 2x}}$$
ADD Note that above gives $$\sum\limits_{k = 0}^{n - 1} {\frac{{\sin \left( {k + \frac{1}{2}} \right)t}}{{\sin \frac{t}{2}}}}  = {\left( {\frac{{\sin \frac{{nt}}{2}}}{{\sin \frac{t}{2}}}} \right)^2}$$ which is fundamental when computing the Fejér Kernel as the Cesaro mean of the Dirichlet Kernel.
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