JEE Advance - Mathematics (1998 - No. 24)
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
$${b_0} = 1,\,b = 3$$
$${b_0} = 0,\,{b_1} = n$$
$${b_0} = - 1,\,{b_1} = n$$
$${b_0} = 0,\,{b_1} = {n^2} - 3n + 3$$
Comments (0)
