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Q. Let n be an odd integer if sin ntheta...

Q. Let n be an odd integer if `sin ntheta=sum_(r=0)^n(b_r)sin^rtheta`, for every value of theta then `b_0 and b_1`---

A

`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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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