Prove $\sin x + \sin 3x + \cdots + \sin(2n-1)x = \frac{\sin^2 nx}{\sin x}$ is true

Question

If $x \neq k\pi$ and $n \in \mathbb Z$ then prove $$\sin x + \sin 3x + \cdots + \sin(2n-1)x = \frac{\sin^2 nx}{\sin x}$$

So I stumbled into this problem in my textbook in the complex numbers chapter. It is an optional problem, so the teacher won't explain it to us. I have no idea on how to approach it in the first place. Could anyone please give me a clue on how to begin solving it?

Answer 1

For a complex numbers proof, note the LHS is the imaginary part of $$S=e^{ix}+e^{3ix}+\cdots+e^{(2n-1)ix}.$$ That's a geometric progression: $$S=e^{ix}\frac{e^{2in x}-1}{e^{2ix}-1}=e^{inx}\frac{e^{inx} -e^{-inx}}{e^{i x}-e^{-ix}} =e^{inx}\frac{\sin nx}{\sin x}=(\cos nx+i\sin nx) \frac{\sin nx}{\sin x}.$$

Answer 2

If we make an induction proof then it is enought to prove: $$\sin(2n+1)x = \frac{\sin^2 (n+1)x}{\sin x} -\frac{\sin^2 nx}{\sin x} $$ or

$$\sin(2n+1)x \cdot \sin x= \Big(\sin (n+1)x -\sin nx\Big)\Big(\sin (n+1)x +\sin nx\Big) $$

Now, since: $$\sin (n+1)x -\sin nx = 2\cos {(2n+1)x\over 2}\sin{x\over 2}$$ and $$\sin (n+1)x +\sin nx = 2\sin {(2n+1)x\over 2}\cos{x\over 2}$$

and using $$ \sin2\alpha = 2\sin \alpha\cos \alpha $$ we are done.