Explanation

(b)
Given, $\sin{n\theta = \sum_{r = 0}^{n}{b_{r}\sin^{r}\theta}}$
$\Rightarrow \ \sin{n\theta = b_{0}.\sin^{0}{\theta + b_{1}\sin^{1}{\theta + b_{2}}}}$
$\sin^{2}{\theta + ... + b_{n}\sin^{n}\theta}$
$\Rightarrow \ \sin{n\theta = b_{0} + b_{1}\sin{\theta + ...b_{n}\sin^{n}\theta}}$
$\because\sin{n\theta =^{n}C_{1}\sin{\theta\cos^{n - 1}{\theta -^{n}C_{3}}}}$
$\sin^{3}{\theta\cos^{n - 3}{\theta + ...}}$
$=^{n}C_{1}\sin{\theta{\left( 1 - \sin^{2}\theta \right)^{\frac{n - 1}{2}}}^{\ }{\ ^{- n}C}_{3}}$
$\sin^{3}{\theta\left( 1 - \sin^{2}\theta \right)^{(n - 3)/2} + ...}$
$\therefore\ \ b_{0} = 0$
$b_{1} =$Coefficient of $\sin{\theta =^{n}C_{1} = n}$
$\lbrack\because n - 1,\ n - 3\ are\ all\ even\ integer\rbrack$
Alternate
$\sin{n\theta = b_{0} + b_{1}\sin{\theta + b_{2}\sin^{2}{\theta + ... + b_{n}\sin^{n}\theta}}}$
Put $\theta = 0$, we get $b_{0} = 1$
Again, $\frac{\sin{n\theta}}{\sin\theta} = \sum_{r = 1}^{n}b_{r}\sin^{r - 1}\theta$
Taking limit as $\theta \rightarrow 0$, we get
$\lim_{\theta \rightarrow 0}{\frac{\sin{n\theta}}{\sin\theta} = b_{1} + 0}$
$\Rightarrow \ \ \ \ \ \ \ n = b_{1}$