Question

已知sin(α+β)sin(α-β)=-1/3

求sin^2α+cos^2β

Answer 1

由sin(α+β)sin(α-β)=-1/3 得
sin(α+β)sin(α-β)=(sinαcosβ+cosαsinβ)(sinαcosβ-cosαsinβ)
=(sinαcosβ)^2-(cosαsinβ)^2
=(sinαcosβ)^2+(sinαsinβ)^2-(sinαsinβ)^2-(cosαsinβ)^2
=sin^2α*(cos^2β+sin^2β)-sin^2β*(sin^2α+cos^2α)
=sin^2α-sin^2β=-1/3
所以
sin^2α+cos^2β=sin^2α+1-sin^2β
=1-1/3
=2/3 。

Answer 2

已知-1/3=sin(α+β)sin(α-β)……积化和差公式：sinθsinφ = -[cos(θ+φ)-cos(θ-φ)]/2

=-[cos2a-cos2β]/2……倍角公式：cos2a=1-2sin^2a；cos2β=2cos^2β-1

=-[(1-2sin^2a)-(2cos^2β-1)]/2
=-[2-2(sin^2a+cos^2β)]/2
=-1+(sin^2a+cos^2β)
sin^2a+cos^2β=2/3

Answer 3

2/3

将上式的左边展开为
sin^2a*cos^2b-sin^2b*cos^2a=-1/3;
将sin^2a变成1-cos^2a展开合并同类项得
cos^2b-cos^2a=-1/3;
变一下cos^2a即得
sin^2a+cos^2b=2/3

相信我，没错的！

Answer 4

a+b=A,a-b=B.a=(A+B)/2,b=(A-B)/2.
后式=1/2(COS(A-B)-COS(A+B))+1
=2/3
最后结果口算不知对不对但方法是这样