Show that $\sin(2x) + \sin(4x) + \ldots+ \sin(2nx) = \dfrac{\sin(nx)\sin((n+1)x)}{\sin(x)}$
I tried to use induction.
Base case is easy, but I'm stuck at the induction step (from $k$ to $k+1$). So I have:
$$
\sin(2x) + \sin(4x) + \ldots + \sin(2kx) + \sin(2(k+1)x)\\
= \frac{\sin(kx)\sin((k+1)x) + \sin(x)\sin((k+2)x)}{\sin(x)}
$$
Where do I go from here?
Best Answer
Umm....wouldn't it be easier to use complex numbers. Take $z=e^{i2x}$ and add the series $z+z^2+z^3+...+z^{2(k+1)x}$, which is a G.P. series hence find the sum and equate the imaginary part....