Let `z=costheta+isin theta`, where `i=sqrt(-1)`. Then the value of `sum_(m=1)^(15)Im(z^(2m-1))` at `theta=2^(@)` is

Options

A. `1/(sin2^(@))`
B. `1/(3sin2^(@))`
C. `1/(2sin2^(@))`
D. `1/(4sin2^(@)`

Answer

D

Explanation

We have,
`z=costheta+isintheta rArr z^(2m-1)=cos(2m-1)theta+isin(2m-1)theta`
`therefore sum_(m=1)^(15)"Im"(z^(2m-1))=sum_(m=1)^(15)sin(2m-1)theta`
`rArr sum_(m=1)^(15)"Im"(z^(2m-1)) = (sin(theta-29theta)/(2)sin15theta)/(sintheta)=(sin^(2)15theta)/(sin theta)`
At `theta=2^(@)`, we have
`sum_(m=1)^(15)(z^(2m-1)) = (sin^(2)30^(@))/(sin2^(@))=(sin^(2)30^(@))/(sin2^(@)) = 1/(4sin2^(@))`