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)]_数学解答

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)]

人气：367 ℃　时间：2019-10-28 05:03:12

解答

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)...