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| 1. |
Sin3x + sin3(2π/3 + x) + sin3(4π/3 + x) = -3/4sin3x |
| Answer» 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,\xa0\\(=> 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 | |