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| 1. |
If sin +cos=a prove sin6+cos6=4-3(a²-1)²/4 |
| Answer» LHS{tex} = {\\sin ^6}\\theta + {\\cos ^6}\\theta {/tex}{tex} = {\\left( {{{\\sin }^2}\\theta } \\right)^3} + {\\left( {{{\\cos }^2}\\theta } \\right)^3}{/tex}{tex} = {\\left( {{{\\sin }^2}\\theta + {{\\cos }^2}\\theta } \\right)^3} - 3{\\sin ^2}\\theta {\\cos ^2}\\theta \\left( {{{\\sin }^2}\\theta + {{\\cos }^2}\\theta } \\right){/tex}{tex}\\left[ {\\because {a^3} + {b^3} = {{(a + b)}^3} - 3ab(a + b)} \\right]{/tex}{tex} = {(1)^3} - 3{\\sin ^2}\\theta {\\cos ^2}\\theta \\times 1{/tex}\xa0{tex}\\left[ {\\because {{\\sin }^2}\\theta + {{\\cos }^2}\\theta = 1} \\right]{/tex}{tex} = 1 - 3{\\sin ^2}\\theta {\\cos ^2}\\theta {/tex}RHS\xa0{tex} = \\frac{{4 - 3{{({x^2} - 1)}^2}}}{4}{/tex}{tex} = \\frac{{4 - 3{{\\left\\{ {{{\\left( {\\sin \\theta + \\cos \\theta } \\right)}^2} - 1} \\right\\}}^2}}}{4}{/tex}\xa0{tex}\\left[ {given\\;x = \\sin \\theta + \\cos \\theta } \\right]{/tex}{tex} = \\frac{{4 - 3{{\\left\\{ {{{\\sin }^2}\\theta + {{\\cos }^2}\\theta + 2\\sin \\theta \\cos \\theta - 1} \\right\\}}^2}}}{4}{/tex}{tex} = \\frac{{4 - 3{{\\left\\{ {1 + 2\\sin \\theta \\cos \\theta - 1} \\right\\}}^2}}}{4}{/tex}\xa0{tex}\\left[ {\\because {{\\sin }^2}\\theta + {{\\cos }^2}\\theta = 1} \\right]{/tex}{tex} = \\frac{{4 - 3{{\\left( {2\\sin \\theta \\cos \\theta } \\right)}^2}}}{4}{/tex}{tex} = \\frac{{4 - 3 \\times 4{{\\sin }^2}\\theta {{\\cos }^2}\\theta }}{4}{/tex}{tex} = \\frac{{4\\left( {1 - 3{{\\sin }^2}\\theta {{\\cos }^2}\\theta } \\right)}}{4}{/tex}LHS = RHSHence proved. | |