Question

已知cos(π/2-α)=根号2cos(3/2π+β),根号3sin(3π/2-α)=-根号2sin(π/2+β),且0

Answer 1

cos(π/2-α)=根号2cos(3/2π+β)-->sin(α)=根号2*cos(3/2π+β) =根号2*cos(1/2π-β)=根号2*sin(β)

3sin(3π/2-α)=-根号2sin(π/2+β)-->根号3*sin(3π/2-α)= -根号3*sin(π/2+α)=-根号2*sin(π/2+β)-->根号3*sin(π/2+α)=根号2*sin(π/2+β)-->根号3*sin(π/2+α)=根号3*cos(-α)= 根号2*sin(π/2+β)= 根号2*cos(-β)--〉根号3*cos(α)=根号2*cos(β)

由此知：sin(α)=根号2*sin(β),根号3*cos(α)=根号2*cos(β),联立方程式,消去α

[sin(β)]^2=1/4,sin(β)=1/2( 0＜β＜π,sin(β)>0)

β=π/6,或5π/6

当β=π/6时,α=π/4

当β=5π/6时,α=3π/4