已知函数f(x)=cosx-cos(x+π/2) x∈R（1）求f（x）的最大值和最小值（2）求f(x)的单调递增区间

f(x)=cosx-cos(x+π/2)
=-2(sin(x+x+π/2)/2sin(x-x-π/2)/2
=-2(sin(x+π/4)sin(-π/4)
=√2sin(x+π/4)
所以最大值是√2,最小值是-√2
增区间 2kπ+π/2>=x+π/4>=2kπ-π/2
2kπ+π4>=x>=2kπ-3π/4区间为增区间
减区间 2kπ+3π/2>=x+π/4>=2kπ+π/2
2kπ+5π/4>=x>=2kπ+π/4区间为减区间