Question

cos(π/2-α)=根号2cos(3/2π+β),根号3sin(3π/2-α)=-根号2sin(π/2+β)且0<α,β<π,求α和β的值

Answer 1

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