Question

sinxcosy=1/2,求cosx-siny范围. 答案：[－3/2,1/2] 求详细过程.

Answer 1

取x=-y=π/4,
则sinx=cosy=√2/2,满足sinxcosy=1/2
此时罩蚂型,cosx=√2/2,siny=-√2/2,
cosx-siny=√2,
√2并不在你的答案：[－3/2,1/2]范围内.
所以,答案显然不正物帆确
应该是：
设物猜m=cosx-siny
m^2=(cosx)^2+(siny)^2-2cosxsiny
=2-(sinx)^2-(cosy)^2±2√[1-(sinx)^2][1-(cosy)^2]
=2-(sinx)^2-(cosy)^2±2√[1-(sinx)^2-(cosy)^2+(sinxcosy)^2]
=2-(sinx)^2-(cosy)^2±√[5-4(sinx)^2-4(cosy)^2]
设t=√[5-4(sinx)^2-4(cosy)^2]