A proof by induction and trigonometry Do you know how to prove that $\displaystyle\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ using induction?

I have tried with $n = 1$ which gives $\cos \frac{x}{2} = \frac{\sin(nx)}{(2\sin1/2x)}$ 
I am not sure on how to expand with the trigonometric formulas.
With $n= p+1$ I get LHS: $\cos(2(n+1)-1)$ which I summaries to $\cos(2n+1)$ which should be $\cos 2n \cos 1-\sin 2n \sin1$ plus the RHS $\frac{\sin(nx)}{(2\sin 1/2x)}$
RHS $p+ 1 = \frac{\sin(n+1x)}{(2\sin 1/2x)}$
Any ideas on how to proceed would be very helpful.

If $$\sum_{r=1}^n\cos\dfrac{(2r-1)}2x=\dfrac{\sin nx}{2\sin\dfrac x2}$$ holds true for $n=m$
$$\sum_{r=1}^{m+1}\cos\dfrac{(2r-1)}2=\dfrac{\sin mx}{2\sin\dfrac x2}+\cos\dfrac{[2(m+1)-1]}2x$$
$$=\dfrac{\sin mx+2\sin\dfrac x2\cos\dfrac{(2m+1)}2x}{2\sin\dfrac x2}$$
Using Werner's formula, $2\sin\dfrac x2\cos\dfrac{(2m+1)}2x=\sin(m+1)x-\sin mx$
