cosx+sinx的单调性怎么算(要有过程)
5个回答

cosx+sinx

=√2(√2/2cosx+√2/2sinx)

=√2(sinπ/4cosx+cosπ/4sinx)

=√2sin(x+π/4)

当2kπ-π/2≤x+π/4≤2kπ+π/2时,即2kπ-3π/4≤x≤2kπ+π/4时,√2sin(x+π/4)单调增

当2kπ+π/2≤x+π/4≤2kπ+3π/2时,即2kπ+π/4≤x≤2kπ+7π/4时,√2sin(x+π/4)单调减

所以

当2kπ-3π/4≤x≤2kπ+π/4时,cosx+sinx单调增

当2kπ+π/4≤x≤2kπ+7π/4时,cosx+sinx单调减