Question

若cosα+cosβ+cosγ=0,sinα+sinβ+sinγ=0,求cos(α-β)的值

怎么做的,过程!

Answer 1

cosα+cosβ=-cosγ
sinα+sinβ=-sinγ
cos^2α+cos^2β+2cosαcosβ=cos^2γ
sin^2α+sin^2β+2sinαsinβ=sin^2γ
相加：2+2(cosαcosβ+sinαsinβ)=1
cosαcosβ+sinαsinβ=-1/2
cos(α-β)=-1/2

Answer 2

sinγ=-sina-sinb
cosγ=-cosa-cosb
∴sin²γ+cos²γ=(-sina-sinb)²+(-cosa-cosb)²=(sin²a+cos²a)+(sin²b+cos²b)+2(cosacosb+sinasinb)
=1+1+2cos(a-b)=1
∴cos(a-b)=-1/2