Prove by induction: $\sin{x}+\sin{3x}+\dots+\sin{(2n-1)x}=\frac{\sin^2{nx}}{\sin{x}}$

Question

Prove by induction: $$\sin{x}+\sin{3x}+\dots+\sin{(2n-1)x}=\frac{\sin^2{nx}}{\sin{x}}$$

I tried the problem using the normal rule of induction(the first principle), but I failed.I failed to make the form $\sin^2{(m+1)x}$. Somebody help me.

Answer 1

HINT

Note that by the product to sum property, we have that

$$\sin(x)\sin((2k-1)x)=\frac12\left(\cos((2k-2)x)-\cos(2kx)\right)$$ This implies that $$\sin((2k-1)x)=\frac{\frac12\left(\cos((2k-2)x)-\cos(2kx)\right)}{\sin x}$$ Use $\cos 2a=1-2 \sin^2 a$ for all $a$. Can you take it from here?

Answer 2

You know that $$ \sin y=\frac{e^{iy}-e^{-iy}}{2i} $$ so you have $$ \frac{\sin^2nx}{\sin x }+\sin(2n+1)x= \left(\frac{e^{inx}-e^{-inx}}{2i}\right)^2\frac{2i}{e^{ix}-e^{-ix}}+ \frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{2i} $$ This becomes $$ \frac{e^{2inx}-2+e^{-2inx}+(e^{i(2n+1)x}-e^{-i(2n+1)x})(e^{ix}-e^{-ix})} {2i(e^{ix}-e^{-ix})} $$ The numerator is $$ e^{2inx}-2+e^{-2inx}+e^{i(2n+2)x}-e^{-2inx}-e^{2inx}+e^{-i(2n+2)x}= (e^{i(n+1)x}-e^{-i(n+1)x})^2 $$ so finally $$ \frac{\sin^2nx}{\sin x }+\sin(2n+1)x= \frac{(e^{i(n+1)x}-e^{-i(n+1)x})^2}{2i(e^{ix}-e^{-ix})}= \frac{(2i)^2\sin^2(n+1)x}{2i\cdot 2i\sin x} $$

Answer 3

For $n=1$ it's obvious.

Thus, it remains to prove that $$\frac{\sin^2nx}{\sin{x}}+\sin(2n+1)x=\frac{\sin^2(n+1)x}{\sin{x}}$$ or $$1-\cos2nx+\cos2nx-\cos(2n+2)x=1-\cos(2n+2)x,$$ which is obvious.