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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
${b_0} = 1,{b_1} = 3$
${b_0} = 0,{b_1} = n$
${b_0} = - 1,{b_1} = n$
${b_0} = 0,{b_1} = {n^2} - 3n + 3$
Solution
(b) Given $\sin \,n\theta = \sum\limits_{r = 0}^n {{b_r}{{\sin }^r}\theta } $
==> $\sin n\theta = {b_0}{\sin ^0}\theta + {b_1}\sin {\,^1}\theta $
$ + {b_2}{\sin ^2}\theta + {b_3}{\sin ^3}\theta + ….. + {b_n}{\sin ^n}\theta $
==> $\sin n\theta = {b_0} + {b_1}\sin \theta + {b_2}{\sin ^2}\theta + …. + {b_n}{\sin ^n}\theta $
($n$ is an odd integer)
$ = {\,^n}{C_1}\sin \theta .{(1 – {\sin ^2}\theta )^{(n – 1)/2}}$
$ – {\,^n}{C_3}{\sin ^3}\theta {(1 – {\sin ^2}\theta )^{(n – 3)/2}} + ….$
$\therefore \,\,\,{b_0} = 0,{b_1} = $ coefficient of $\sin \theta = {\,^n}{C_1} = n$
($ n -1= n -3$ are all even integers)
