How to find out coefficient $b_1$ in summation $\sin n\theta=\sum_{r=0}^{n} b_r \sin^r\theta$ 
Let $n$ be a odd integer. If 
  $$\sin  n\theta=\sum_{r=0}^{n} b_r \sin^r\theta$$
  for every value of $\theta$. I have to find $b_1$.

I don't know how to start this?
Thanks.
 A: It's $n$. First, consider this complex identity:
$$e^{in\theta}=(\cos\theta+i\sin\theta)^n$$
$$e^{in\theta}=\cos(n\theta)+i\sin(n\theta)$$
By the binomial development:
$$(\cos\theta+i\sin\theta)^n=\cos^n\theta+in\cos^{n-1}\theta\sin\theta+(powers\,of\,\sin\theta>1)=$$
Because $n$ odd and $n-1$ even:
$$\cos^n\theta+in(\cos^2\theta)^{(n-1)/2}\sin\theta+(powers\,of\,\sin\theta>1)=$$
$$\cos^n\theta+in(1-\sin^2\theta)^{(n-1)/2}\sin\theta+(powers\,of\,\sin\theta>1)=$$
$$\cos^n\theta+in\sin\theta+(powers\,of\,\sin\theta>1)$$
Now:
$$\cos(n\theta)+i\sin(n\theta)=\cos^n\theta+in\sin\theta+(powers\,of\,\sin\theta>1)$$
Taking the imaginary part:
$$\sin(n\theta)=n\sin\theta+ (powers\,of\,\sin\theta>1)$$
And it's done with the bonus of $b_0=0$
A: $$\sin  n\theta=\sum_{r=0}^{n} b_r \sin^r\theta=b_0+b_1\sin \theta+\sum_{r=2}^{n} b_r \sin^r\theta$$
by differentiation of this equivalency we have
$$n\cos n\theta=b_1\cos \theta+\sum_{r=2}^{n} b_r r\sin^{r-1}\theta\cos\theta$$
set $\theta=2\pi$ you take
$$\color{blue}{b_1=n}$$
