求函数y=sinA-3/cosA-2的值域为

步骤如下：
1.ycosA-2y=sinA-3
2.ycosA-sinA=2y-3
sinA-ycosA=3-2y
3.令r^2=1^2+(-y)^2=1+y^2,r>0
cosB=1/r,sinB=-y/r
4.3-2y=sinA-ycosA
=r(1/r*sinA-y/r*cosA)
=r(sinAcosB+cosAsinB)
=rsin(A+B)
5.sin(A+B)=(3-2y)/r
因为|sinx|≤1,
6.|3-2y|/r≤1 (3-2y)^2≤r^2=1+y^2
9-12y+3y^2≤0 y^2-4y+3≤0 (y-1)(y-3)≤0
1≤y≦3
故：函数y=sinA-3/cosA-2的值域为[1,3]