What is `[(sin (pi)/6 +i(1-cos (pi/6)))/(sin (pi)/6 -(1-cos (pi/6)))]^3` where `i=sqrt(-1)` equal to

Options

A. 1
B. `-1`
C. `i`
D. `-i`

Answer

C

Explanation

`[(sin""(pi)/(6)+i(1-cos""(pi)/(6)))/(sin""(pi)/(6)-i(1-cos""(pi)/(6)))]^(3)`
`=[(2sin""(pi)/(12)cos""(pi)/(12)+i(2sin^(2)""(pi)/(12)))/(2sin""(pi)/(12)cos""(pi)/(12)-i(2sin^(2)""(pi)/(12)))]^(3)`
`=[(cos""(pi)/(12)+isin""(pi)/(12))/(cos""(pi)/(12)-isin""(pi)/(12))]^(3)=((e^(i)""(pi)/(12))/(e^(-i)""(pi)/(12)))^(3)`
`=(e^(i(pi)/(6)))^(3)=e^(ixx3xx(pi)/(6))=e^(i(pi)/(2))=cos""(pi)/(2)+1sin""(pi)/(2)=1`
`therefore` (c) is correct.