Prove by induction that

$$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ with $n \geq 1$

  1. Testing n=1:

$$\sin(x)= \frac {1-\cos(2x)}{2\sin(x)}$$




$$\sin^2(x)+\cos^2(x) =1$$

It shows that n=1 yields a true identity (Pythagorean identity)

  1. Let's assume that $P_n$ is true: $$\ sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$

  2. Let's consider adding $\sin(2n+1)x$ to $P_n$:

$$ \sin(x) +\sin(3x) +...+ \sin (2n-1)x+ \sin(2n+1)x= \frac{1-\cos(2nx)}{2\sin x} + \sin(2n+1)x$$

Considering the RHS: $$\frac{1-\cos(2nx)}{2\sin x} + \sin (2n+1)x$$ $$\frac{1-\cos(2nx)+ \sin(2n+1)x \cdot 2\sin x}{2 \sin x}$$ $$\frac{1-\cos(2nx)+ 2[\sin(2n+1)x \cdot sin x]}{2sinx}$$ $$\frac{1-\cos(2nx)+ 2 \cdot \frac{1}{2} [\cos[(2n+1)x-x]-\cos[(2n+1)x+x]]}{2\sin x}$$ $$\frac{1-\cos(2nx)+ \cos(2nx)-\cos(2n+2)x}{2\sin x}$$ $$\frac{1-\cos(2n+2)x}{2\sin x}$$

It follows that $$\sin(x) +\sin(3x) +...+ \sin [(2(n+1)-1)x]= \frac{1-\cos(2(n+1)x)}{2\sin x}$$

Therefore, $$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ is true

Any input is much appreciated.


Also, you can multiply both sides by $2\sin x$ and apply telescoping:

$$2\sin x\sin x +2\sin 3x\sin x +…+ 2\sin (2n-1)x\sin x=$$ $$[\cos0-\require{cancel}\cancel{\cos2x}]+[\cancel{\cos2x}-\cancel{\cos4x}]+\cdots+[\cancel{\cos(2n-2)x}-\cos2nx]=$$ $$1-\cos2nx$$

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You can also use $$\sin(kx)=\frac{e^{ikx}-e^{-ikx}}{2i}$$and geometric series for a direct proof.

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  • $\begingroup$ I have better. Just use $\sin(x)=\Im(e^{ix})$. This is much simpler. $\endgroup$ – Simply Beautiful Art May 27 '17 at 13:05

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