telescoping sum

Question

I had to use the identity $\sin^{4}(x)=\sin^{2}(x)-\frac{1}{4}\sin^{2}(2x)$ to write the sum $\sum^\infty_{k=1} 4^k \sin^{4}(\frac{\theta}{2^k})$ as a telescoping sum. However, how would I use an identity for $\sin(3x)$ to figure out a similar telescoping sum problem. I am wondering if someone would help me out with that. I believe I would not have a problem to solve a sum, but I would not be able to come up with one.

Answer 1

\begin{align} & \overbrace{\sin(3x) = \sin (2x+x)}^\text{Get used to this sort of thing!} \\[6pt] = {} & \sin(a+b) \\[6pt] = {} & \sin a\cos b+\cos a\sin b \\[6pt] = {} & \sin(2x)\cos x + \cos(2x)\sin x \end{align} Then use the double-angle formulas for sine and cosine.