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