Question #71dff

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Answer 1 · 2017-01-26T08:37:31

Ist part of the expression

$=3(sinx-cosx)^4$

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

$=3(1-2sinxcosx)^2$

$=3-12sinxcosx+12sin^2xcos^2x$

2nd part of the expression

$=6(sinx+cosx)^2$

$=6(sin^2x+cos^2x+2sinxcosx)$

$=6+12sinxcosx$

3rd part of the expression

$=4(sin^6x+cos^6x)$

$=4((sin^2x+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x)$

$=4(1-3sin^2xcos^2x)$

$=4-12sin^2xcos^2x$

So whole part

$= 3 -12sinxcosx+12sin^2xcos^2x +6+12sinxcosx+4-12sin^2xcos^2x$

$=13$