1) 2sinxcosx=cos2x-2sin^2x 2)cos^2a-sin^2a=2cos^2a*tga/tg2a - вопрос №507489012222222222 от artemnovik506 10.10.2020 17:29

Question

Алгебра: 1) 2sinxcosx=cos2x-2sin^2x 2)cos^2a-sin^2a=2cos^2a*tga/tg2a

artemnovik506 · Answer

1)2sinxcosx=cos2x-2sin^2xsin2x=cos2x-(1+cos2x)sin2x= -12x= -pi/2+2pi*n  n∈Zx= -pi/4+pi*n  n∈Z2)sin^2a=2cos^2a*tqa/tq2a(1-cos2a)/2=(2cosa*sina)/tq2a(1-cos2a)/2=sin2a/tq2a(1-cos2a)/2=cos2a1-c0s2a=2cos2acos2a=12a=pi*n  n∈Za=(pi*n)/2    n∈Z...