Question

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

• 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 $0=b_0$
$n\theta ={\sum }_{r=1}^n{b}_r\sin {}^r\theta$
$\Rightarrow \frac{\sin n\theta }{\sin \theta }={\sum }_{r=1}^n{b}_r{\left(\sin \theta \right)}^{r-1}$
$=b_1+b_2\sin \theta +b_3{\sin }^2\theta +.....+b_n{\sin }^{n-1}\theta$
Taking limit as $\theta \to 0$, we obtain
${\lim }_{\theta \to 0}\frac{\sin n\theta }{\sin \theta }=b_1\Rightarrow b_1=n.$
$\left[\because {\lim }_{\theta \to 0}\frac{\sin n\theta }{\theta }={\lim }_{\theta \to 0}\frac{n\cos n\theta }{\cos \theta }=\frac{n\times 1}{1}=n\right]$