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Question Number 111426 by weltr last updated on 03/Sep/20

Answered by Dwaipayan Shikari last updated on 03/Sep/20

4(sin^6 α+cos^6 α)−3cos^2 2α  =4(sin^2 α+cos^2 α)^3 −12sin^2 αcos^2 α−3cos^2 2α  =4−3sin^2 2α−3cos^2 2α  =4−3=1

4(sin6α+cos6α)3cos22α=4(sin2α+cos2α)312sin2αcos2α3cos22α=43sin22α3cos22α=43=1

Commented by weltr last updated on 03/Sep/20

thank you

thankyou

Answered by bobhans last updated on 03/Sep/20

(sin^2 x)^3 +(cos^2 x)^3 =1.(sin^4 x−(sin xcos x)^2 +cos^4 x)   = 1−2(sin xcos x)^2 −(sin xcos x)^2    = 1−(3/4)(2sin xcos x)^2 = 1−(3/4)sin^2  2x  (∗) 4{1−(3/4)sin^2 2x}−3cos^2 3x          = 4−3sin^2 2x−3cos^2 2x = 4−3(1)=1

(sin2x)3+(cos2x)3=1.(sin4x(sinxcosx)2+cos4x)=12(sinxcosx)2(sinxcosx)2=134(2sinxcosx)2=134sin22x()4{134sin22x}3cos23x=43sin22x3cos22x=43(1)=1

Commented by weltr last updated on 03/Sep/20

thank you

thankyou

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