作业帮f（X）=（1+根号下3倍的tanX）/[1+(tanX)平方],求单调递增区间
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/01 22:21:39
![f（X）=（1+根号下3倍的tanX）/[1+(tanX)平方],求单调递增区间](/uploads/image/z/10373998-22-8.jpg?t=f%EF%BC%88X%EF%BC%89%3D%EF%BC%881%2B%E6%A0%B9%E5%8F%B7%E4%B8%8B3%E5%80%8D%E7%9A%84tanX%EF%BC%89%2F%5B1%2B%28tanX%29%E5%B9%B3%E6%96%B9%5D%2C%E6%B1%82%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
f（X）=（1+根号下3倍的tanX）/[1+(tanX)平方],求单调递增区间
f（X）=（1+根号下3倍的tanX）/[1+(tanX)平方],求单调递增区间
f（X）=（1+根号下3倍的tanX）/[1+(tanX)平方],求单调递增区间
f(x) =(1+√3tanx)/(1+tan^2x).
f(x)=1+√3tanx)/sec^2x.
=(1+√3tanx)*cos^2x.
=cos^2x+√3sinxcosx.
=(1+cos2x)/2+√3/2*sin2x.
=(1/2)cos2x+√3/2sin2x+1/2.
=sin2xcos30+sin30cos2x+1/2.
∴f(x)=sin(2x+π/6)+1/2.
∵ 2kπ-π/2