Let n be an odd integer. If sin ntheta = suml | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\sin n	heta = \sum\limits_{r = 0}^n b_r \sin^r 	heta$. Since $\sin n	heta$ is an odd function of $	heta$ when $n$ is odd,the expansion

Vedclass · Accepted Answer

$b_0 = 0, b_1 = n$

Vedclass · Answer

(B) Given $\sin n	heta = \sum\limits_{r = 0}^n b_r \sin^r 	heta$. Since $\sin n	heta$ is an odd function of $	heta$ when $n$ is odd,the expansion in terms of $\sin 	heta$ must contain only odd powers of $\sin 	heta$. Thus,$b_0 = 0$ and $b_2 = b_4 = \dots = 0$. Using the expansion $\sin n	heta = 	ext{Im}((\cos 	heta + i \sin 	heta)^n)$,we have $\sin n	heta = \binom{n}{1} \sin 	heta \cos^{n-1} 	heta - \binom{n}{3} \sin^3 	heta \cos^{n-3} 	heta + \dots$. Substituting $\cos^2 	heta = 1 - \sin^2 	heta$,we get $\sin n	heta = \binom{n}{1} \sin 	heta (1 - \sin^2 	heta)^{(n-1)/2} - \binom{n}{3} \sin^3 	heta (1 - \sin^2 	heta)^{(n-3)/2} + \dots$. The coefficient of $\sin 	heta$ is obtained from the first term: $\binom{n}{1} \sin 	heta (1)^ {(n-1)/2} = n \sin 	heta$. Therefore,$b_1 = n$ and $b_0 = 0$.

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

Similar Questions

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

If $\cos \theta = \frac{-3}{5}$ and $\theta$ does not lie in the second quadrant,then $\tan \frac{\theta}{2} = $

For the value of $\frac{2 \tan(x)}{1-\tan^2(x)}$ to be positive,find the values of $x$ such that $x \in \left(0, \frac{\pi}{2}\right)$.

$\frac{\sec 8\theta - 1}{\sec 4\theta - 1}$ is equal to

If $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module