已知cos2α-cos2β=1/4,那么sin(α+β)sin(α-β)=?
2个回答

cos^2α-cos^2β=1/4……1式

1-sin^2α-1+sin^2β=1/4

sin^2β-sin^2α=1/4……2式

1式+2式,提

cos^2α-cos^2β+sin^2β-sin^2α=1/2

(cos^2α-sin^2α)-(cos^2β-sin^2β)=1/2

cos2α-cos2β=1/2

cos[(α+β)+(α-β)]-cos[(α+β)-(α-β)]

=[cos(α+β)cos(α-β)-sin(α+β)sin(α-β)]-[cos(α+β)cos(α-β)+sin(α+β)sin(α-β)]

=-2sin(α+β)sin(α-β)

=1/2

-2sin(α+β)sin(α-β)=1/2

所以

sin(α+β)sin(α-β)=-1/4

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