Prove that `sintheta+sin3theta+sin5theta+.....+sin(2n-1)theta=(sin^2ntheta)/(sintheta)dot`

Explanation

Here, we will use,

`sinalpha+sin(alpha+beta)+sin(alpha+2beta)+...+sin(alpha+(n-1)beta) = (sin((nbeta)/2))/(sin(beta/2))**sin(alpha+((n-1)beta)/2)`

Now,

`L.H.S. = sintheta+sin3theta+sin5theta+...+sin(2n-1)theta`

`=sintheta+sin(theta+2theta)+sin(theta+2*2theta)+...sin(theta+(n-1)2theta)`

`=(sin((2ntheta)/2))/(sin((2theta)/2))**sin(theta+((n-1)2theta)/2)`

`=(sin(ntheta)/(sintheta))*sin(theta+ntheta-theta)`

`=sin^2(ntheta)/sintheta = R.H.S.`