Question

函数f(x)=sinx+cosx的单调递增区间是__

Answer 1

f(x)=sinx+cosx
=√2(√2/2*sinx+√2/2cosx)
=√2(sinxcosπ/4+cosxsinπ/4)
=√2sin(x+π/4)
递增则2kπ-π/2<x+π/4<2kπ+π/2
2kπ-3π/4<x<2kπ+π/4
所以增区间(2kπ-3π/4,2kπ+π/4)

Answer 2

f(x)=½sin(2x),-π/2+2kπ≤2x≤π/2+2kπ,解得：-π/4+kπ≤x≤π/4+kπ，
所以单调递增区间为[-π/4+kπ,π/4+kπ]
k∈Z