Trigonometrical Equations 1 Question 13
13. Let $n$ be an odd integer. If $\sin n \theta=\sum _{r=0}^{n} b _r \sin ^{r} \theta$, for every value of $\theta$, then
(1998, 2M)
(a) $b _0=1, b _1=3$
(b) $b _0=0, b _1=n$
(c) $b _0=-1, b _1=n$
(d) $b _0=0, b _1=n^{2}-3 n+3$
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Answer:
Correct Answer: 13. (c)
Solution:
- Given, $\sin n \theta=\sum _{r=0}^{n} b _r \sin ^{r} \theta$
Now, put $\quad \theta=0$, we get $0=b _0$
$\therefore \quad \sin n \theta=\sum _{r=1}^{n} b _r \sin ^{r} \theta$
$\Rightarrow \quad \frac{\sin n \theta}{\sin \theta}=\sum _{r=1}^{n} b _r(\sin \theta)^{r-1}$
Taking limit as $\theta \to 0$
$$ \Rightarrow \lim _{\theta \to 0} \frac{\sin n \theta}{\sin \theta} = \lim _{\theta \to 0} \sum _{r=1}^{n} b_r (\sin \theta)^{r-1} $$
$$ \Rightarrow \lim _{\theta \to 0} \frac{n \theta \cdot \frac{\sin n \theta}{n \theta}}{\theta \cdot \frac{\sin \theta}{\theta}} = b_1 + 0 + 0 + 0 + \ldots $$
$[\because$ other values becomes zero for higher powers of $\sin \theta$ ]
$$ \begin{aligned} \Rightarrow & \frac{n \cdot 1}{1} & =b _1 \\ \Rightarrow & b _1 & =n \end{aligned} $$
