求函数的周期和单调区间y=(3根号2)/2[sin(x/2-π/12)+cos(x/2-π/12)]

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 03:25:06
求函数的周期和单调区间y=(3根号2)/2[sin(x/2-π/12)+cos(x/2-π/12)]

求函数的周期和单调区间y=(3根号2)/2[sin(x/2-π/12)+cos(x/2-π/12)]
求函数的周期和单调区间
y=(3根号2)/2[sin(x/2-π/12)+cos(x/2-π/12)]

求函数的周期和单调区间y=(3根号2)/2[sin(x/2-π/12)+cos(x/2-π/12)]
y=(3根号2)/2[根号2sin(x/2-π/12+π/4)]
=(3根号2)/2[根号2sin(x/2+π/6)]
所以周期T=4π
令g(x)=sin(x/2+π/6)
所以g(x)在(4kπ-4/3π,4kπ+2/3π)增
(4kπ+2/3π,4kπ+8/3π)减
因为f(x)=(3根号2)/2根号2g(x)
所以f(x)在(4kπ-4/3π,4kπ+2/3π)减
(4kπ+2/3π,4kπ+8/3π)增