Let n be odd integer . If `sin theta = sum_(r=0)^(n) b_(r) sin^(r) theta`, for every value of `theta`, thenA. `b_(0) = 1, b_(1) =3`B. `b_(0) = 0, b_(1) `C. `b_(0) = - 1 , b_(1) =n `D. `b_(0) =0, b_(1) = n^(2) - 3n +3`

Let n be odd integer . If `sin theta = sum_(r=0)^(n) b_(r) sin^(r) the

1.	Let n be odd integer . If `sin theta = sum_(r=0)^(n) b_(r) sin^(r) theta`, for every value of `theta`, thenA. `b_(0) = 1, b_(1) =3`B. `b_(0) = 0, b_(1) `C. `b_(0) = - 1 , b_(1) =n `D. `b_(0) =0, b_(1) = n^(2) - 3n +3`
Answer» Correct Answer - B Given, `sin n theta = underset(r=0)overset(n)sum b_(r) sin^(r) theta` Now put, `theta = 0`, we get `0 =b_(0)` `therefore sin ntheta =underset(r=1)overset(n) b_(r) sin^(r) theta` `rArr (sin n theta)/(sin theta) = underset(r=1)overset(n) sum b_(r) (sin theta)^(r-1)` Taking limit as ` theta to0` `rArr underset( theta to 0)(lim) (sin ntheta)/(sin theta) = underset(theta to 0)(lim) underset(r=1)overset(n)sum b_(r) (sin theta)^(n-1)` `rArr underset(theta to 0)(lim) (n theta.(sin ntheta)/(n theta))/(theta.(sintheta)/(theta)) = b_(1) + 0 + 0 + 0.....` [`because` other values become zero for higher power of sin `theta`] `rArr " "(n.1)/(1) =b_(1)` `rArr " "b_(1) = n`