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EKymanK
@EKymanK
November 2021
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dasdasfa
94) 1) cos^2( α+2β) + sin^2 (α-2β)-1=c0s^2 (α+2β) + sin^2 (α-2β)-sin^2 (α-2β)- cos^2 (α-2β)=cos^2 (α+2β) -cos^2 (α-2β)=
=(cos(α+2β) -cos(α-2β) ) *(cos(α+2β) +cos(α-2β))=
-2sinα sin2β * 2cosα cos2β=-sin2α *sin4β;
95) 1)(cos4a -cos2a)/(sin3a sina)=(-2sin3a sina) /(sin3a sina)=-2;
96) 1)=(4sin^2 a- (2sin a cosa)^2 )/( 4*(sin^2 a+cos^2 a)-4sin^2 a -4sin^2 a cos^2 a) )=(4sin^2 a(1-cos^2 a)) / (4sin^2 a+4cos^2 a -4sin^2 a -4sin^2 a cos^2 a)=(4sin^2 a sin^2 a) /(4cos^2 a (1-sin^2 a))=4sin^4 a / (4cos^4 a)=
=tg^4 a
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Answers & Comments
=(cos(α+2β) -cos(α-2β) ) *(cos(α+2β) +cos(α-2β))=
-2sinα sin2β * 2cosα cos2β=-sin2α *sin4β;
95) 1)(cos4a -cos2a)/(sin3a sina)=(-2sin3a sina) /(sin3a sina)=-2;
96) 1)=(4sin^2 a- (2sin a cosa)^2 )/( 4*(sin^2 a+cos^2 a)-4sin^2 a -4sin^2 a cos^2 a) )=(4sin^2 a(1-cos^2 a)) / (4sin^2 a+4cos^2 a -4sin^2 a -4sin^2 a cos^2 a)=(4sin^2 a sin^2 a) /(4cos^2 a (1-sin^2 a))=4sin^4 a / (4cos^4 a)=
=tg^4 a