sinα=1/5,α∈(0,π/2),
=>cosα=√[1-sin²α]=√[1-(1/5)^2]=(2/5)√6cosβ=3/4,β∈(3π/2,2π)=> sinβ=-(1/4)√7,cosβ=3/4
sin(α-β)=sinαcosβ-cosαsinβ
=(1/5)*(3/4)+[(1/4)√7]*[(2/5)√6]=3/20+(2/20)√42=(3+2√42)/20
sinα=1/5,α∈(0,π/2),
=>cosα=√[1-sin²α]=√[1-(1/5)^2]=(2/5)√6cosβ=3/4,β∈(3π/2,2π)=> sinβ=-(1/4)√7,cosβ=3/4
sin(α-β)=sinαcosβ-cosαsinβ
=(1/5)*(3/4)+[(1/4)√7]*[(2/5)√6]=3/20+(2/20)√42=(3+2√42)/20