What is the real part of `(sin x +icos x)^3` where `i=sqrt(-1)?`

Options

A. `-cos 3x`
B. `-sin 3x`
C. `sin 3x`
D. `cos 3x`

Answer

B

Explanation

`(sinx+i cosx)^(3)`
`=sin^(3)x+(i)^(3)cos^(3)x+3i(sinx)(xcos)(sinx+icosx)`
`=sin^(3)x-icos^(3)x+3isin^(2)xcosx-3sinxcos^(2)x`
`=sin^(3)x-3sinx cos^(2)x-icosx(cos^(2)x+sin^(2)x)`
`=sinx(sin^(2)x-3cos^(2)x)-icosx(cos^(2)x+3sin^(2)x)`
Real part of `(sin x+icosx)^(3)`
`=sin^(3)x(sin^(2)x-3cos^(2)x)`
`=sin x[sin^(2)x-3(1-sin^(2)x)]`
`=sinx[4sin^(2)x-3]`
`=4sin^(3)x-3sinx`
`=-(3sinx-4sin^(3)x)=-sin3x`
`therefore` Option (b) is correct.