Explanation

(b)
Given, $\sin{n\ \theta = \sum_{r = 0}^{n}{b_{r}\sin^{r}{\theta = b_{0} + b_{1}\sin\theta + b_{2}\sin^{2}\theta + ... + b_{n}\sin^{n}\theta}\ \ldots(i)}}$
Putting $\theta = 0\$in Eq.(i), we get$\ 0 = b_{0}$
Again, Eq. (i)can be written as sin $n\ \theta = \sum_{r = 0}^{n}{b_{r}\sin^{r}\theta}$
$\frac{\sin{n\theta}}{\sin\theta} = \sum_{r = 1}^{n}{b_{r}\sin^{r - 1}\theta}$
Taking limit as $\theta \rightarrow 0,\$we get $\lim_{\theta \rightarrow 0}{\frac{\sin{n\theta}}{\sin\theta} = b_{1}}$
$\Rightarrow \ \lim_{\theta \rightarrow 0}{n\left( \frac{\sin{n\theta}}{n\theta} \right)\left( \frac{\theta}{\sin\theta} \right) = b_{1}}$
$\Rightarrow n = b_{1}\$Hence, $b_{0 =}0;\ b_{1} = n$