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


*

*Testing n=1:

$$\sin(x)= \frac {1-\cos(2x)}{2\sin(x)}$$
$$2\sin^2(x)=1-\cos(2x)$$
$$2\sin^2(x)=1-[\cos^2(x)-\sin^2(x)]$$
$$\sin^2(x)=1-\cos^2(x)$$
$$\sin^2(x)+\cos^2(x) =1$$
It shows that n=1 yields a true identity (Pythagorean identity)


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


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