Explanation

(b)
Given, $\sin{n\theta = \sum_{r = 0}^{n}b_{r}}\sin^{r}\theta$
$\Rightarrow \ \sin{n\theta = b_{0}\sin^{0}\theta + b_{1}\sin^{1}\theta + b_{2}\sin^{2}\theta + b_{3}\sin^{3}\theta + ... + b_{n}\sin^{n}\theta}$
$\Rightarrow \ \sin{n\theta = b_{0} + b_{1}\sin\theta + b_{2}\sin^{2}\theta + ... + b_{n}\sin^{n}\theta}$
($n$ is an odd integer)
$\because\sin{n\theta = \ ^{n}C_{1}\sin{\theta\cos^{n - 1}}\theta - \ ^{n}C_{3}\sin^{3}\theta\cos^{n - 3}\theta + ...}\$
$= \ ^{n}C_{1}sin\theta{(1 - \sin^{2}\theta)}^{\frac{(n - 1)}{2}} - \ ^{n}C_{3}\sin^{3}\theta{(1 - \sin^{2}\theta)}^{\frac{(n - 3)}{2}} + ...$
$\therefore b_{0} = 0,\ b_{1} = coefficient\ of\ \sin{\theta = \ ^{n}C_{1} = n}$
$(\because n - 1,n - 3$ are all even integers)