cos(π/2-α)=根号2cos(3/2π+β),根号3sin(3π/2-α)=-根号2sin(π/2+β)且0
问题描述：
cos(π/2-α)=根号2cos(3/2π+β),根号3sin(3π/2-α)=-根号2sin(π/2+β)且0
答
cos(π/2-α)=根号2cos(3/2π+β),
sinα=√2sinβ--------->sin^2α=2sin^2β--------------(1)
根号3sin(3π/2-α)=-根号2sin(π/2+β)
-√3cosα=-√2cosβ------>3cos^2α=2cos^2β---------(2)
(1)+(2)
sin^2α+3cos^2α=2sin^2β+2cos^2β
1+2cos^2α=2 ,cos^2α=1/2 ,α=π/4
sinα=√2sinβ ,sinβ=1/2,β=π/6