Question

Prove that $\cos x+\cos 3x+.......+\cos (2n-1)x=\frac{\sin 2nx}{2\sin x},x\ne K\pi ,K\in I$ and then deduce than $\sin x+3\sin 3x+......+(2n-1)\sin (2n-1)x=\frac{\left[\right(2n+1\left)\sin \right(2n-1)x-(2n-1\left)\sin \right(2n+1\left)x\right]}{4{\sin }^{2}x}$

Answer 1

$\cos x+\cos 3x+.....\cos (2n-1)x=\frac{\sin 2nx}{2\sin x}$

This is true for $n=1,n=k$

Now,for $n=k+1$

add $\cos \left(2\right(k+1)-1)x\Rightarrow \cos (2k+1)$ on both sides which is next term in series.

$\therefore \cos x+....\cos (2k+1)x=\frac{\sin 2kx}{2\sin x}+\cos (2k+1)xLHS=\frac{\sin 2kx+2\sin x\cos (2k+1)x}{2\sin x}=\frac{\sin 2kx+2\sin x[\cos 2kx\cos x-\sin 2kx\sin x]}{2\sin x}=\frac{\sin 2kx+2\sin x\cos x\cos 2kx-2\sin {}^{2}x\sin 2kx}{2\sin x}=\frac{\sin 2kx(1-2\sin {}^{2}x)+\sin 2x}{2\sin x}=\frac{\left(\frac{\sin 2kx}{\cos 2x}\right)x+\left(\frac{\sin 2x}{\cos 2kx}\right)x}{2\sin x}=\frac{\sin (2kx+2x)}{2\sin x}=\frac{\sin 2x(k+1)}{2\sin x}=RHS$