Proving a sum using induction I am having a problem with this question.
I need to prove by induction that: $$\sum_{k=1}^n \sin(kx)=\frac{\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}$$
The relation is obvious for n=1
Now I suppose that the relation is true for a natural number n and I want to show that $$\sum_{k=1}^{n+1} \sin(kx)=\frac{\sin(\frac{n+2}{2}x)\sin(\frac{n+1}{2}x)}{\sin(\frac{x}{2})}$$
We have $$\sum_{k=1}^{n+1} \sin(kx)=\sin[(n+1)x]+\sum_{k=1}^{n} \sin(kx)
                                 =\sin[(n+1)x]+\frac{\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}= \frac{\sin[(n+1)x]\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}$$
I am unable to simplify the expression using the trigonometric identities. I keep turning around in circles.
Can somebody help me please.
Thank you in advance

You can actually use telescopy, which is just induction in disguise.
We have that
$$\cos b - \cos a = 2\sin \frac{{a + b}}{2}\sin \frac{{a - b}}{2}$$
Now let $$b=\left(k+\frac 1 2 \right)x$$
$$b=\left(k-\frac 1 2 \right)x$$
Then
$$\cos \left( {k + \frac{1}{2}} \right)x - \cos \left( {k - \frac{1}{2}} \right)x = 2\sin kx\sin \frac{x}{2}$$
Now sum through $k=1,\dots,n$, to get
$$\sum\limits_{k = 1}^n {\cos \left( {k + \frac{1}{2}} \right)x - \cos \left( {k - \frac{1}{2}} \right)x}  = 2\sin \frac{x}{2}\sum\limits_{k = 1}^n {\sin kx} $$
$$\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2} = 2\sin \frac{x}{2}\sum\limits_{k = 1}^n {\sin kx} $$
whence 
$$\frac{{\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2}}}{{2\sin \frac{x}{2}}} = \sum\limits_{k = 1}^n {\sin kx} $$
But using our first formula once more, we have
$$\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2} = 2\sin \frac{{\left( {n + 1} \right)x}}{2}\sin \frac{{nx}}{2}$$
so finally
$$\frac{{\sin \frac{{\left( {n + 1} \right)x}}{2}\sin \frac{{nx}}{2}}}{{\sin \frac{x}{2}}} = \sum\limits_{k = 1}^n {\sin kx} $$
as desired. 
