Question

sin^2a+sin^2b-sin^2a * sin^2b+cos^2a * cos^2b 化简？

Answer 1

结果是1
(sina)^2+(sinb)^2-(sina)^2*(sinb)^2+(cosa)^2*(co *** )^2
=(sina)^2*(1-sinb^2)+(sinb)^2+(cosa)^2*(co *** )^2
=sina^2*co *** ^2+cosa^2*co *** ^2+sinb^2
=(sina^2+cosa^2)*co *** ^2+sinb^2
=co *** ^2+sinb^2
=1,2,原式=sin^2a(1-sin^2b)+sin^2b+cos^2a*(1-sin^2b)=(1-sin^2b)(sin^2a+cos^2a)+sin^2b=cos^2b+sin^2b=1,2,=sin^2(a)*(1-sin^2(b))+sin^2(b)+cos^2(a)*cos^2(b)
=(sin^2(a)+cons^2(a))*cos^2(b)+sin^2(b)
=cos^2(a)+sin^2(b)
=1,2,1
sin^2a(1-sin^2b)+sin^2b+cos^2acos^b
=sin^2a*cos^2b+sin^2b+cos^2acos^2b
=cos^2b*(sin^2a+cos^2a)+sin^2b
=cos^2b+sin^2b
=1,1,=1,1,1+1=2,1,sin^2a+sin^2b-sin^2a * sin^2b+cos^2a * cos^2b
=sin^2a-sin^2a * sin^2b+cos^2a * cos^2b+sin^2b
=sin^2a(1- sin^2b)+cos^2a * cos^2b+sin^2b
=sin^2a cos^2b+cos^2a * cos^2b+sin^2b
= cos^2b(sin^2a +cos^2a )+sin^2b
=cos^2b+sin^2b
=1,0,