1. The value of $$3(sin x- cos x)^4$$ + $$6(sin x + cos x)^2$$ + $$4(sin^{6}x + cos^{6}x)$$ is
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By: anil on 05 May 2019 01.58 am
Expression : $$3(sin x- cos x)^4$$ + $$6(sin x + cos x)^2$$ + $$4(sin^{6}x + cos^{6}x)$$ = $$3 (sin x - cos x)^2 (sin x - cos x)^2 + 6 (sin^2 x + cos^2 x + 2 . sin x . cos x) + 4 [(sin^2 x)^3 + (cos^2 x)^3]$$ = $$3 (sin^2 x + cos^2 x - 2 . sin x . cos x) ( sin^2 x + cos^2 x - 2 . sin x . cos x) + 6 (1 + 2 . sin x . cos x) + 4[(sin^2 x + cos^2 x) (sin^4 x + cos^4 x - sin^2 x . cos^2 x)]$$ = $$3 (1 - 2 . sin x . cos x)^2 + 6 + 12 . sin x . cos x + 4 [(sin^2 x + cos^2 x)^2 - 3 . sin^2 x . cos^2 x]$$ = $$3 (1 + 4 . sin^2 x . cos^2 x - 4 . sin x . cos x) + 6 + 12 . sin x . cos x + 4 - 12 . sin^2 x . cos^2 x$$ = $$3 + 12 . sin^2 x . cos^2 x - 12 . sin x . cos x + 6 + 12 . sin x . cos x + 4 - 12 . sin^2 x . cos^2 x$$ = $$3 + 6 + 4 = 13$$
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