cos(α-β)=cosαcosβ+sinαsinβ
(sinα+sinβ)^2=(sinα)^2+(sinβ)^2+2sinαsinβ
所以2sinαsinβ=9/25-(sinα)^2-(sinβ)^2
同理2cosαcosβ=16/25-(cosα)^2-(cosβ)^2
cos(α-β)=1/2(2cosαcosβ+2sinαsinβ)
=1/2[9/25-(sinα)^2-(sinβ)^2+16/25-(cosα)^2-(cosβ)^2]
=1/2{(9/25+16/25)-[(sinα)^2+(cosα)^2]-[(sinβ)^2+(cosβ)^2]}
因为(sinα)^2+(cosα)^2=1,(sinβ)^2+(cosβ)^2=1
所以=1/2[1-1-1]=-1/2
(^2代表平方） cos(α-β)=cosαcosβ+sinαsinβ
(sinα+sinβ)^2=(sinα)^2+(sinβ)^2+2sinαsinβ
所以2sinαsinβ=9/25-(sinα)^2-(sinβ)^2
同理2cosαcosβ=16/25-(cosα)^2-(cosβ)^2
cos(α-β)=1/2(2cosαcosβ+2sinαsinβ)
=1/2[9/25-(sinα)^2-(sinβ)^2+16/25-(cosα)^2-(cosβ)^2]
=1/2{(9/25+16/25)-[(sinα)^2+(cosα)^2]-[(sinβ)^2+(cosβ)^2]}
因为(sinα)^2+(cosα)^2=1,(sinβ)^2+(cosβ)^2=1
所以=1/2[1-1-1]=-1/2
(^2代表平方）