1.

Prove that `sin^3alpha + sin^3(120^@ +alpha)+ sin^3(240^@+ alpha)=-3/4sin3alpha`A. `3/4sin3alpha`B. `-3/4sin3 alpha`C. `4/3sin3 alpha`D. `-4/3sin3 alpha`

Answer» Correct Answer - B
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`


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