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  • Let n be an odd integer. If  <img alt="\sin n \theta=\sum_{r=0}^n b_r \sin ^{r} \theta" src="https://entrancecorn

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

 

Option: 1

b_0=1, b_1=3


Option: 2

b_0=0, b_1=n


Option: 3

b_0=-1, b_1=n


Option: 4

b_0=0, b_1=n^2-3 n+3


Answers (1)

best_answer

Given  \sin n \theta=\sum_{r=0}^n b_r \sin ^r \theta \\

\Rightarrow \sin n \theta=b_0 \sin ^0 \theta+b_1 \sin ^1 \theta +b_2 \sin ^2 \theta+b_3 \sin ^3 \theta+\ldots . .+b_n \sin ^{n} \theta \\

\Rightarrow \sin n \theta=b_0+b_1 \sin \theta+b_2 \sin ^2 \theta+\ldots .+b_n \sin ^n \theta \\

                                                                              (n is an odd integer)

Using the formula of \text { \ } \sin n \theta

\text { \ } \sin n \theta={ }^n C_1 \sin \theta \cos ^{n-1} \theta-{ }^n C_3 \sin ^3 \theta \cos ^{n-3} \theta+\ldots . \\

={ }^n C_1 \sin \theta \cdot\left(1-\sin ^2 \theta\right)^{(n-1) / 2} -{ }^n C_3 \sin ^3 \theta\left(1-\sin ^2 \theta\right)^{(n-3) / 2}+\ldots . \\

\therefore b_0=0, b_1=\text { coefficient of } \sin \theta={ }^n C_1=n \\

                                  ( n-1=n-3  are all even integers) }

Posted by

seema garhwal

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