Question

已知sin α -cosβ=1/3,sinβ+cosα =1/2

则sin(β- α)=?

请写出过程，谢谢

Answer 1

因为 sin α -cosβ=1/3,sinβ+cosα =1/2
所以 (sin α -cosβ)^2 = 1/9 , (sinβ+cosα)^2 = 1/4
所以 (sin α -cosβ)^2 + (sinβ+cosα)^2 = 1/9 + 1/4
即 (sin α)^2 - 2sin αcosβ + (cosβ)^2 + (sinβ)^2 + 2sinβcosα + (cosα)^2 = 13/36
所以 2sinβcosα - 2sin αcosβ + (sin α)^2 + (cosα)^2 + (cosβ)^2 + (sinβ)^2 = 13/36
所以 2sin(β- α) = 13/36 - 2
得到 sin(β- α) = -59/72

Answer 2

sin(β- α)=sinβ * cosα - cosβ * sin α
sin α -cosβ=1/3 两边平方
sinβ+cosα =1/2 两边平方
把两个式子相加得到 sinβ * cosα - cosβ * sin α = 13/36 = sin(β- α)

Answer 3

⑴sin α -cosβ=1/3,⑵sinβ+cosα =1/2
⑴平方+⑵平方，得2+2sin(β- α)=13/36
sin(β- α)= --59/72

Answer 4

sin(β- α)=sinαcosβ-cosαsinβ