Question

函数f(x)=sin(4x+3分之派）+cos(4x-6分之派）的最小正周期和递减区间

Answer 1

f(x)=sin(4x+π/3）+cos(4x-π/6)
=sin(4x+π-π/6）+cos(4x-π/6)
=sin(π+4x-π/6）+cos(4x-π/6)
=-sin(4x-π/6）+cos(4x-π/6)
=-√2[√2/2*sin(4x-π/6）+√2/2*cos(4x-π/6)]
=-√2[sin(4x-π/6)cosπ/4+cos(4x-π/6)sinπ/4]
=-√2sin(4x-π/6+π/4)
=-√2sin(4x+π/12)
T=2π/4=π/2
递减区间
2kπ+π/2<4x+π/12<2kπ+3π/2
2kπ+5π/12<4x<2kπ+17π/12
kπ/2+5π/48<x<kπ/2+17π/48

Answer 2

周期2分之π，递减区间：（24分之π+2分之kπ，24分之7π+2分之kπ）

Answer 3

先分别展开，然后再化简合并得2sin（4x+3分之派），剩下的你应该可以求出了吧