Question

Let n be an odd integer. If $\sin \, n \theta = \sum^n_{r=0} \, b_r \sin^r \theta$, for every value of $\theta$, then

• A. $b_0 = 1, b_1 = 3$
• B. $b_0 = 0, b_1 = n$
• C. $b_0 = - 1, b_1 = n$
• D. $b_0 = 0, b_1 = n^2 - 3n + 3 $

Answer 1

Answer: (B)

Putting $\theta = 0$, we get $0 = b_0$
$n \theta = \sum^n_{r = 1} \, b_r \sin {^r\theta}$
$\Rightarrow \frac{\sin n \theta}{\sin\theta} = \sum^{n}_{r=1} b_{r} \left(\sin\theta\right)^{r-1} $
$= b_{1}+ b_{2} \sin\theta + b_{3} \sin^{2}\theta + .....+ b_{n} \sin^{n-1} \theta$
Taking limit as $\theta \to 0$, we obtain
$\lim_{\theta \to0} \frac{\sin n \theta}{\sin\theta} = b_{1} \Rightarrow b_{1} = n. $
$ \left[\because \lim_{\theta\to0} \frac{\sin n \theta}{\theta} = \lim_{\theta \to0} \frac{n \cos n \theta}{\cos\theta} = \frac{n \times1}{1} = n \right]$