1-1/4sin^2 2A-sin^2B-cos^4A
问题描述：
1-1/4sin^2 2A-sin^2B-cos^4A
sin(A+B)=3/5
sin(A-B)=-4/5
答
1-1/4sin^2 2A-sin^2B-cos^4A
=1-sin^2Acos^2A-sin^2B-cos^4A
=1-cos^2A-sin^2B
sin(A+B)=sinAcosB+cosAsinB=3/5
sin(A-B)=sinAcosB-cosAsinB=-4/5
cosAsinB=7/10
sinAcosB=-1/10
sin^2Acos^2B=(1-cos^2A)(1-sin^2B)
=1-sin^2B-cos^2A+cos^2Asin^2B
=1-sin^2B-cos^2A+49/100
=1/100
1-sin^2B-cos^2A=-12/25
所以：1-1/4sin^2 2A-sin^2B-cos^4A=-12/25.