Question #77e3a

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Question #77e3a

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Answer 1 · 2016-12-21T05:44:50

$L H S = 2 {sin}^{6} x + 2 {cos}^{6} x + 1$ $LHS=2sin^6x+2cos^6x+1$

$= 2 ({({sin}^{2} x)}^{3} + {({cos}^{2} x)}^{3} + 1$ $=2((sin^2x)^3+(cos^2x)^3+1$

$= 2 ({({sin}^{2} + {cos}^{2} x)}^{3} - 3 {sin}^{2} x {cos}^{2} x ({sin}^{2} x + {cos}^{2} x) + 1$ $=2((sin^2+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x)+1$

$= 2 \cdot {(1)}^{3} - 3 \cdot 2 \cdot {sin}^{2} x {cos}^{2} x \cdot (1) + 1$ $=2*(1)^3-3*2*sin^2xcos^2x*(1)+1$

$= 3 - 3 \cdot 2 \cdot {sin}^{2} x {cos}^{2} x$ $=3-3*2*sin^2xcos^2x$

$= 3 (1^{2} - 2 \cdot {sin}^{2} x {cos}^{2} x)$ $=3(1^2-2*sin^2xcos^2x)$

$= 3 ({({sin}^{2} x + {cos}^{2} x)}^{2} - 2 \cdot {sin}^{2} x {cos}^{2} x)$ $=3((sin^2x+cos^2x)^2-2*sin^2xcos^2x)$

$= 3 {({({sin}^{2} x)}^{2} + ({cos}^{2} x))}^{2}$ $=3((sin^2x)^2+(cos^2x))^2$

$= 3 ({sin}^{4} x + {cos}^{4} x)$ $=3(sin^4x+cos^4x)$

$= 3 {sin}^{4} x + 3 {cos}^{4} x = R H S$ $=3sin^4x+3cos^4x=RHS$

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Question #77e3a

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