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माना $n$ एक विषम पूर्णांक है। यदि $\theta $ के सभी मानों के लिये $\sin n\theta = \sum\limits_{r = 0}^n {{b_r}{{\sin }^r}\theta } $ हो, तो
${b_0} = 1,{b_1} = 3$
${b_0} = 0,{b_1} = n$
${b_0} = - 1,{b_1} = n$
${b_0} = 0,{b_1} = {n^2} - 3n + 3$
Solution
(b) दिया है $\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 $
($n$ विषम पूर्णांक है)
${ = ^n}{C_1}\sin \theta {(1 – {\sin ^2}\theta )^{(n – 1)/2}}$
$ – {\,^n}{C_3}{\sin ^3}\theta {(1 – {\sin ^2}\theta )^{(n – 3)/2}} + ….$
$\therefore \,\,\,{b_0} = 0,{b_1} = $$\sin \theta $ का गुणांक $ = {\,^n}{C_1} = n$
( $n -1= n -3$ सभी सम पूर्णांक हैं)
