sin(α+β)sin(α-β)=1/3 求证:1/4(sin2α)^2+(sinβ)^2+(cosα)^4 的定值我是照着题抄下来的,一个是证明吧

问题描述:

sin(α+β)sin(α-β)=1/3 求证:1/4(sin2α)^2+(sinβ)^2+(cosα)^4 的定值
我是照着题抄下来的,一个是证明吧

1/4(sin2α)^2+(sinβ)^2+(cosα)^4
=(sinα)^2(cosα)^2+(cosα)^4+(sinβ)^2
=(cosα)^2+(sinβ)^2
我已经有思路了 分我拿定了··
参考这个积化和差公式:
sinα·sinβ=-(1/2)[cos(α+β)-cos(α-β)]
sin(α+β)sin(α-β)=-1/2(cos2α-cos2β)=1/3
cos2α-cos2β=-2/3=1-2(sinβ)^2+1-2(cosα)^2
所以(cosα)^2+(sinβ)^2 =(-2/3-2)/-2=2/3
···无语 才算出来
分给
侯宇诗吧···

2/3

1/4(sin2α)^2+(sinβ)^2+(cosα)^4=(sinαcosα)^2+(sinβ)^2+(cosα)^2*(cosα)^2=(cosα)^2[(sinα)^2+(cosα)^2]+(sinβ)^2=(cosα)^2+(sinβ)^2=Mcos(a-b)-cos(a+b)=2sinasinb2/3=2sin(α+β)sin(α-β)=cos(2...

是证明“是定值”,还是求出该定值