若sinα-sinβ=根号3⼀2,cosα-cosβ=1⼀2,则cos(α-β)的值为

求计算过程。我怎么算都是0啊、
2024-12-31 16:04:30
推荐回答(2个)
回答1:

(sina-sinb)^2+(cosa-cosb)^2
=(sin^2a-2sina*sinb+sin^2b)+(cos^2a-2cosacosb+cos^2b)
=(sin^2a+cos^2a)+(sin^2b+cos^2b)-2(sinasinb+cosacoab)
=2-2cos(a-b)
=(√3/2)²+(1/2)²
=3/4+1/2
=1
即 2-2cos(a-b)=1 所以cos(α-β)=1/2

回答2:

两个式子平方后相加=>sa2+sb2+ca2+cb2-2cacb-2sasb =3/4+1/4
=>1/2=cacb+sasb =>c(a-b)=1/2