若sinα-sinβ=1-根号三/2 cosα-cosβ=-1/2 则 cos(α-β)等于

问题描述:

若sinα-sinβ=1-根号三/2 cosα-cosβ=-1/2 则 cos(α-β)等于

sina-sinb=1-根号3/2
平方:sin^2a+sin^2b-2sinasinb =1-根号3+3/4=7/4-根号3..............1
cosa-cosb=-1/2
平方:cos^2a+cos^2b-2cosacosb=1/4....................2
1式加2式:
1+1-2(cosacosb+sinasinb)=2-根号3
2cos(a-b)=根号3
cos(a-b)=根号3/2

(sinα-sinβ)=1-√3/2(sinα-sinβ)²=(1-√3/2)²sin²α-2sinαsinβ+sin²β=7/4 -√3 (1)cosα-cosβ=-1/2(cosα-cosβ)²=1/4cos²α-2cosαcosβ+cos²β=1/4 ...