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

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(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

$b_0 = 0, b_1 = n$

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(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

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