Sin3x + sin3(2π/3 + x) + …

Ans. we know

$sin^3x = sinx(sin^2x)$

$=>{ sin x (1-cos2x)\over2}$

$=> {1\over 2} [{sin x -sin x cos2x}]$

$=> {1\over 2} [{ sin x - {1\over 2} [sin (x+2x) +sin (x-2x)]}]$

$=> {1\over 2} [{ sin x - {1\over 2} [sin 3x -sin x}]]$

$=> {1\over 2}\times{1\over 2} [{2 sin x - sin 3x +sin x}]$

$=> {3sin x - sin 3x \over4}$  (1)

Similarly,

$=> sin^3({{2\pi\over 3} +x})= {3sin ({{2\pi\over 3 }+x })- sin (2\pi +3x )\over4}$

$=> {3[sin ({{2\pi\over 3 })cos x +sin xcos( {2\pi\over 3} })]- sin 3x\over4}$

$=> {3[{\sqrt3\over 2}cos x -{1\over 2}sin x)]- sin 3x\over4}$

$=> {3\sqrt3cos x -3sin x- 2sin 3x\over 8}$    (2)

Similarly, 

$=> sin^3({{4\pi\over 3} +x})= {3sin ({{4\pi\over 3 }+x })- sin (4\pi +3x )\over4}$

$=> {3[sin ({{4\pi\over 3 })cos x +sin xcos( {4\pi\over 3} })]- sin 3x\over4}$

$=> {3[(-{\sqrt3\over 2 })cos x +sin x({-1\over 2})]- sin 3x\over4}$

$=> {-3\sqrt3cos x -3sin x-2sin 3x\over 8}$  (3)

From (1),(2) and (3)

$=> sin^3{x}+ sin^3({{2\pi\over 3} +x}) + sin^3({{4\pi\over 3} +x})$

$=>{3sin x - sin3x \over 4}+ {3\sqrt3cos x -3sin x- 2sin 3x\over 8} + {-3\sqrt3cos x -3sin x- 2sin 3x\over 8}$

$=>{1\over 8}[{6sin x - 2sin3x + 3\sqrt3cos x -3sin x- 2sin 3x -3\sqrt3cos x -3sin x- 2sin 3x}]$

$=>{1\over 8}[ {- 6sin 3x}] = {-6\over 8} {sin 3x}$

$=> {-3\over 4}{sin 3x} = RHS$

Hence Proved