Prove that: `sin^3A+sin^3((2pi)/3+A)+sin^3((4pi)/3+A)=-3/4sin3Adot`

Options

A. `3/4sin3alpha`
B. `-3/4sin3 alpha`
C. `4/3sin3 alpha`
D. `-4/3sin3 alpha`

Answer

B

Explanation

We have,
`sinalpha+sin((2pi)/(3)+alpha)+sin((4pi)/(3)+alpha)`
`=sin alpha+2 sin(pi+alpha)cos""(pi)/(3)=0`
`thereforesin^(3)alpha+sin^(3)((2pi)/(3)+alpha)+sin^(3)((4pi)/(3)+alpha)`
`=3 sinalpha sin((2pi)/(3)+alpha)+sin^(3)((4pi)/(3)+alpha)`
`=3sinalphasin(120^(@)+alpha)sin(240^(@)+alpha)`
`=-3sinalphasin(60^(@)-alpha)sin(60^(@)-alpha)sin(60^(@)+alpha)=-(3)/(4)sin3 alpha`