y=(sinx+2)(cosx-2)值域怎么求？

y=(sinx+2)(cosx+2)=sinxcosx+2(sinx+cosx)+4
=1/2sin2x+2√2sin(x+π/4)+4
=-1/2cos(2x+π/2)+2√2sin(x+π/4)+4
=-1/2(1-2sin(x+π/4)^2)+2√2sin(x+π/4)+4
=sin(x+π/4)^2+2√2sin(x+π/4)+7/2
=[sin(x+π/4)+√2]^2+3/2
则值域为[9/2-2√2,9/2+2√2]

可以看成(2,-2)和(cosx,sinx)连线的斜率
(cosx,sinx)又可以看成单位圆上的一点
这样斜率就介于两条切线之间