函数y=sin(x+π/2)单调区间
问题描述:
函数y=sin(x+π/2)单调区间
答
y=sin(x+π/2)
=cosx
所以
x∈【2Kπ,2kπ+π】单调递减
x∈【2kπ+π,2Kπ+2π】单调递增.