y=sinx+cosx
y'=cosx-sinx=√2[√2/2cosx-√2/2sinx]=√2cos(x+π/4)
下面求单调递增区间
y'>=0,
2kπ-π/2 ≤x+π/4 ≤ 2kπ+π/2
2kπ-3π/4 ≤x≤ 2kπ+π/4
单调递增区间[2kπ-3π/4 ,2kπ+π/4]
下面求单调递减区间
y'
求y=sinx+cosx的单调区间?
=0,2k"}}}'>
y=sinx+cosx
y'=cosx-sinx=√2[√2/2cosx-√2/2sinx]=√2cos(x+π/4)
下面求单调递增区间
y'>=0,
2kπ-π/2 ≤x+π/4 ≤ 2kπ+π/2
2kπ-3π/4 ≤x≤ 2kπ+π/4
单调递增区间[2kπ-3π/4 ,2kπ+π/4]
下面求单调递减区间
y'