Question

已知sin(α+β)sin(α-β)=1/3.（1）求cos2β-cos2α的值；（2)求(1/4)(sin2α)^2+(sinβ)^2+(cosα)^4

已知sin(α+β)sin(α-β)=1/3.（1）求cos2β-cos2α的值；（2）求(1/4)(sin2α)^2+(sinβ)^2+(cosα)^4的值.

答案：（1）2/3；（2）2/3.

需要详细过程。麻烦各位了。

Answer 1

(1)因为sin(α+β)sin(α-β)=1/3
而cos2β-cos2α
=-2*sin(α+β)sin(β-α)
=2sin(α+β)sin(α-β)
=2/3
(2)(1/4)(sin2α)^2+(sinβ)^2+(cosα)^4
=1/4*(1-cos^2(2α))+(1-cos2β)/2+[(1+cos(2α))/2]^2
=1/4+(1-cos2β)/2+1/4+cos(2α)/2
=1/41/2+1/4-(cos2β-cos2α)/2
=1-1/3
=2/3

Answer 2

(1)cos2β-cos2α=2sin(α+β)sin(α-β)=2/3
(2)(1/4)(sin2α)^2+(sinβ)^2+(cosα)^4=(sinβ)^2+(1/4)(sin2α)^2+(cosα)^4=(1-cos2β)/2+(sin2α/2)^2+(cosα)^2*(cosα)^2=1/2-cos2β/2+(sinαcosα)^2+(cosα)^2*(cosα)^2=1/2-cos2β/2+(cosα)^2*[(sinα)^2+(cosα)^2)=1/2-cos2β/2+(cosα)^2=1/2-cos2β/2+(1+cos2α)/2=1-(cos2β-cos2α)/2=1-1/3=2/3

Answer 3

sin(α+β)sin(α-β)=(sinαcosβ+cosαsinβ)(sinαcosβ-cosαsinβ)
=(sinαcosβ)^2-(cosαsinβ)^2=(1-cos^2α)cos^2β-(1-cos^2β)cos^2α
=cos^2β-cos^2α=1/3
cos2β-cos2α=2cos^2β-1-2cos^2α+1=2cos^2β-2cos^2α=2/3

(1/4)(sin2α)^2+(sinβ)^2+(cosα)^4=sin^2αcos^2α+sin^2β+cos^4α=cos^2α+sin^2β
=cos^2α+sin^2β=cos^2α+1-co^2β=1-(cos^2β-cos^2α)=2/3

Answer 4

（1）因为 sin(α+β)sin(α-β=1/2{cos[(α+β)-(α-β)]-cos[(α+β)+(α-β)]}=1/2[cos2β-cos2α]=1/3
所以cos2β-cos2α=2/3
(2)原式子1/4*（1- cos4α）/2+ 1/2*(1-cos2β)/2+ [(1+cos2α)/2]^2 =1-(cos2β-cos2α)/2
=1-2/3*1/2=2/3