Question #71dff

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Question #71dff

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Answer 1 · 2017-01-26T08:37:31

Ist part of the expression

$= 3 {(sin x - cos x)}^{4}$ $=3(sinx-cosx)^4$

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

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

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

2nd part of the expression

$= 6 {(sin x + cos x)}^{2}$ $=6(sinx+cosx)^2$

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

$= 6 + 12 sin x cos x$ $=6+12sinxcosx$

3rd part of the expression

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

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

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

$= 4 - 12 {sin}^{2} x {cos}^{2} x$ $=4-12sin^2xcos^2x$

So whole part

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

$= 13$ $=13$

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Question #71dff

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