Sums $\sum_{k=1}^n \sin(2k-1)\theta$, $\sum_{k=1}^n \sin^2(2k-1)\theta $ To prove: 
$1.$ $$\sum_{k=1}^n \sin(2k-1)\theta = \frac{\sin^2 n\theta}{\sin \theta}.$$
$2.$ $$\sum_{k=1}^n \sin^2(2k-1)\theta = \frac{n}{2} - \frac{\sin 4n\theta}{4\sin 2\theta}.$$
 A: They become easy (sums of geometric progressions) if you use 
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
and
$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.$$
When $\sin$ and $\cos$ are replaced in this way by rational functions in $e^{ix}$, you can then replace $e^{ix}$ by $z$ and then many trigonometric identities become either polynomial identities or identities between power series. 
If the equality is between rational functions in $\sin(x)$ and $\cos(x)$ then the above substitutions will give you an equality between rational functions. Clearing denominators it becomes an identity between polynomials. This can be checked just by writing the polynomials in the standard form (sums of powers of $z$). This turns the solution into an algorithm.
For example:
$$\cos(2x)=\cos^2(x)-\sin^2(x)$$
It turns into 
$$\frac{z^2+z^{-2}}{2}=\left(\frac{z+z^{-1}}{2}\right)^2+\left(\frac{z-z^{-1}}{2i}\right)^2$$
Opening parenthesis, bringing all to one side, and writing as sum of powers of $z$ we get $$0z^4+0z^3+0z^2+0z+0=0.$$
In this problem you get sums of progressions with hypergeometric terms. The sum of which can also be determined algorithmically, whenever there is a nice sum. 
A: 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.
