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

题目

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

答案

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