求函数的单调区间:(1)y=sin(π/4-3x),(2)f(x)=sinx(sinx-cosx)
问题描述：
求函数的单调区间:(1)y=sin(π/4-3x),(2)f(x)=sinx(sinx-cosx)
答
(1)y=sin(π/4-3x)
递增2kπ-π/22kπ-3π/42kπ/3-π/4
=sin^2x-sinxcosx
=(1-cos2x)/2-sin2x/2
=1-(sin2x+cos2x)/2
=1-根号2sin(2x+π/4)/2
递增2kπ+π/22kπ+π/4kπ+π/8