Okay, I'm working on a really tricky induction problem where I have to show the following (assuming that $\displaystyle \sin{(x/2)}\neq0$):

$\displaystyle \sin{(x)}+2\sin{(2x)}+\ldots +n\sin{(nx)}=\frac{\sin{[(n+1)x]}}{4\sin^2{(x/2)}}-\frac{(n+1)\cos{[(2n+1)(x/2)]}}{2\sin{(x/2)}}$

for all natural numbers $\displaystyle n$.

I've already verified the base case (with the help of ebaines), so now I'm trying to show that the above equation implies this equation:

$\displaystyle \noindent{ \sin{(x)}+2\sin{(2x)}+\ldots +(n+1)\sin{[(n+1)x]}= \frac{\sin{[(n+2)x]}}{4\sin^2{(x/2)}}-\frac{(n+2)\cos{[(2n+3)(x/2)]}}{2\sin{(x/2)}}}$

I almost hesitate to ask this as to me it seems either impossible or a huge amount of work, but perhaps that's not so for those with more experience. Using the first equation we can write the second equation as

$\displaystyle \noindent{\sin{(x)}+2\sin{(2x)}+\ldots +(n+1)\sin{[(n+1)x]}=\frac{\sin{[(n+1)x]}}{4\sin^2{(x/2)}}-\frac{(n+1)\cos{[(2n+1)(x/2)]}}{2\sin{(x/2)}}+(n+1)\sin{[(n+1)x]}}$

so what we desire to show is that

$\displaystyle \noindent{\frac{\sin{[(n+1)x]}}{4\sin^2{(x/2)}}-\frac{(n+1)\cos{[(2n+1)(x/2)]}}{2\sin{(x/2)}}+(n+1)\sin{[(n+1)x]}=\frac{\sin{[(n+2)x]}}{4\sin^2{(x/2)}}-\frac{(n+2)\cos{[(2n+3)(x/2)]}}{2\sin{(x/2)}}}$

So I'm unsure where to go from here. We already have the denominators set up neatly, so I don't want to mess with that. I thought about distributing the $\displaystyle n+1$ in the third term and then adding the resulting two terms to the first two terms in some manner, but I'm not sure how the arguments will work out. Can anyone help?