解,假设存在,利用(sinα)^2+(cosα)^2=1,由已知sin(3π-α)=sinα=(根号2)cos(π/2-β)=√2sin β,即sinα=√2sin β (1) 根号3cos(-α)=√3cosα=-根号2cos(π+β)=√2cosβ,即)√3cosα=√2cosβ (2)
(1)^2+(2)^2
( sinα)^2+3(cosα)^2=2 2(cosα)^2=1 a∈(0,π),cosα=±√2/2
当cosα=√2/2 a=π/4,带入(1) √2sin β =√2/2 β∈(-π,0),sin β