求函数y=sinx+cosx的单调区间

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'<=0
2kπ+π/2 ≤x+π/4 ≤ 2kπ+3π/2
2kπ+π/4 ≤x≤ 2kπ+4π/3
单调递减区间[2kπ+π/4 , 2kπ+4π/3]

导数=cosx-sinx
令cosx-sinx=0
也就是tanx=0
x=K(Pai/2)

在简单判断一下单调性就好了

y=(√2/2)×sin(x+45)这样判断不是更快吗