f(x)=|sinx|+|cosx|
①x∈[2kπ,2kπ+π/2)时
f(x)=sinx+cosx=√2sin(x+π/4)
x+π/4∈[2kπ+π/4,2kπ+3π/4)
所以 f(x)在[2kπ,2kπ+π/4]上是单调递增的
在(2kπ+π/4,2kπ+π/2)上是单调递减的
当 x=2kπ+π/2时 f(x)=1
②x∈[2kπ+π/2,2kπ+π)时
f(x)=sinx-cosx=√2sin(x-π/4)
x-π/4∈[2kπ+π/4,2kπ+3π/4)
所以 f(x)在[2kπ+π/2,2kπ+3π/4)上是单调递增的
在[2kπ+3π/4,2kπ+π)上是单调递减的
f(2kπ+π)=1
③x∈[2kπ+π,2kπ+3π/2)时
f(x)=-sinx-cosx=-√2sin(x+π/4)
x+π/4∈[2kπ+5π/4,2kπ+7π/4)
所以 f(x)在[2kπ+π,2kπ+5π/4)上是单调递增的
在[2kπ+5π/4,2kπ+3π/2)上是单调递减的
f(2kπ+3π/2)=1
④x∈[2kπ+3π/2,2kπ+2π)
f(x)=-sinx+cosx=-√2sin(x-π/4)
x-π/4∈[2kπ+5π/4,2kπ+7π/4)
所以 f(x)在[2kπ+3π/2,2kπ+7π/4)上是单调递增的
在[2kπ+7π/4,2kπ+2π)上是单调递减的
f(2kπ+2π)=1