Question #77e3a

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Answer 1 · 2016-12-21T05:44:50

$LHS=2sin^6x+2cos^6x+1$

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

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

$=2*(1)^3-3*2*sin^2xcos^2x*(1)+1$

$=3-3*2*sin^2xcos^2x$

$=3(1^2-2*sin^2xcos^2x)$

$=3((sin^2x+cos^2x)^2-2*sin^2xcos^2x)$

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

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

$=3sin^4x+3cos^4x=RHS$

Proved