已知sin(α+β)sin(α-β)=3分之1 求cos2β-cos2α的值,4分之1 sin2α方+sinβ方+cos

解cos2β-cos2α=cos[(a+β)+(β-α)]-cos[(β+α)-(β-α)]=cos(a+β)cos(β-α)-sin(a+β)sin(β-α)-cos(a+β)cos(β-α)-sin(a+β)sin(β-α)
=-2sin(α+β)sin(β-α)
=2sin(α+β)sin(α-β)
=2/3
(sin2α)^2/4+(sinβ)^2+(cosα)^4
=(sin2α)^2/4+(1-cos2β)/2+(1/2+cos2α/2)^2
=(sin2a)^2/4+(1-cos2β)/2+1/4+（cos2α)^2/4
+cos2α/2
=1+1/4+1/2-(cos2β-cos2α)/2
=7/4-(1/3)
=17/12