已知sinα+sinβ=1/2,cosα+cosβ=1/3,求[cos（α-β）/2]的平方的值。

Question

急用！

Answer 1

sinα+sinβ=1/2 =>(sina+sinβ)^2=1/4 <=>sin^2a+2sinasinβ+sin^2β=1/4. (1)
cosα+cosβ=1/3 =>(cosa+cosβ)^2=1/9 <=>cos^2a+2cosacosβ+cos^2β=1/9 (2)
(1)+(2) =>2+2(cosacosβ+sinasinβ)=1/4+1/9=13/36.
<=>2(cosacosβ+sinasinβ)=-59/36
<=>cos(a-β)=-59/72.

[cos(a-β)/2]^2
=[1+cos(a-β)]/2.
=(1-59/72)/2
=(13/72)/2
=13/144.