Question
Let $$n$$ be an odd integer. If $$\sin n\theta = \sum\limits_{r = 0}^n {{b_r}{{\sin }^r}\theta } $$ for all real $$\theta $$ then
A.
$${b_0} = 1,{b_1} = 3$$
B.
$${b_0} = 0,{b_1} = n$$
C.
$${b_0} = - 1,{b_1} = n$$
D.
$${b_0} = 0,{b_1} = {n^2} - 3n - 3$$
Answer :
$${b_0} = 0,{b_1} = n$$
Solution :
$$\sin n\theta = {b_0} + {b_1}\sin \theta + {b_2}{\sin ^2}\theta + .....\,\,\,.$$
This is possible when $$n$$ is an odd integer.
Put $$\theta = 0$$ to get $${b_0}.$$ After differentiating w.r.t. $$\theta ,$$ put $$\theta = 0$$ to get $${b_1}.$$