三角模板函数使用示例

前言

模板函数

  • 核心的模板函数(y=sinx),其性质如下:

定义域:(xin R)

值 域:(y=sinxin [-1,1])

单调性:单增区间 ([2kpi-cfrac{pi}{2},2kpi+cfrac{pi}{2}](kin Z)); 单减区间 ([2kpi+cfrac{pi}{2},2kpi+cfrac{3pi}{2}](kin Z))

奇偶性:奇函数;(sin(-x)=-sinx)

周期性:(T=2pi)

对称性:对称轴(x=kpi+cfrac{pi}{2}(kin Z));对称中心((kpi,0)(kin Z))

零 点:(x=kpi (kin Z))

最 值:(x=2kpi+cfrac{pi}{2})时,(y_{max}=1)(x=2kpi-cfrac{pi}{2})时,(y_{min}=-1)

五点法作图:

(x) (0) (cfrac{pi}{2}) (pi) (cfrac{3pi}{2}) (2pi)
(f(x)=sinx) (0) (1) (0) (-1) (0)
(点的坐标) ((0,0)) ((cfrac{pi}{2},1)) ((pi,0)) ((cfrac{3pi}{2},-1)) ((2pi,0))

效果图如下:

使用示例

当研究清楚了上述的函数(y=sinx)的性质后,我们就能够以此为依托,研究更复杂的正弦型函数的各种性质了。

我们以(y=2sin(2x+cfrac{pi}{6})+1)为例子加以说明;

定义域(xin R)

值域:由于(-1leqslant sin(2x+cfrac{pi}{6})leqslant 1)

((-1) imes 2+1leqslant 2sin(2x+cfrac{pi}{6})+1leqslant 1 imes 2+1),即(-1leqslant yleqslant 3)

单调性:由于(2)倍和后边的(+1)不影响单调性,故利用(y=sin(2x+cfrac{pi}{6}))求单调区间;

(2kpi-cfrac{pi}{2}leqslant 2x+cfrac{pi}{6}leqslant 2kpi+cfrac{pi}{2})(kin Z)

解得单调递增区间为([kpi-cfrac{pi}{3},kpi+cfrac{pi}{6}])((kin Z))

(2kpi+cfrac{pi}{2}leqslant 2x+cfrac{pi}{6}leqslant 2kpi+cfrac{3pi}{2})(kin Z)

解得单调递减区间为([kpi+cfrac{pi}{6},kpi+cfrac{2pi}{3}])((kin Z))

奇偶性:由于(f(0) eq 0),且(f(0))没有取到最值,故函数没有奇偶性;

周期性(T=cfrac{2pi}{2}=pi)

对称性:比如求对称轴方程,此时后边的(+1)不影响其对称性,前边的2倍也不影响,

故利用(y=sin(2x+cfrac{pi}{6}))求对称轴方程,

(2x+cfrac{pi}{6}=kpi+cfrac{pi}{2}(kin Z)),解得对称轴方程为:(x=cfrac{kpi}{2}+cfrac{pi}{6}(kin Z))

求对称中心,先利用(y=sin(2x+cfrac{pi}{6}))求对称中心,最后补充(+1)

(2x+cfrac{pi}{6}=kpi(kin Z)),解得(x=cfrac{kpi}{2}-cfrac{pi}{12}(kin Z))

故对称中心坐标为((cfrac{kpi}{2}-cfrac{pi}{12},1)(kin Z))

零点:令(y=2sin(2x+cfrac{pi}{6})+1=0),即(sin(2x+cfrac{pi}{6})=-cfrac{1}{2})

(2x+cfrac{pi}{6}=2kpi-cfrac{pi}{6}(kin Z))(2x+cfrac{pi}{6}=2kpi-cfrac{5pi}{6}(kin Z))

(x=kpi-cfrac{pi}{6}(kin Z))(x=kpi-cfrac{pi}{2}(kin Z))

最值(y_{min}=-1)(y_{max}=3)

五点法作图:自定周期的起止点;函数为(f(x)=2sin(2x+cfrac{pi}{6})+1)

(x) (-cfrac{pi}{12}) (cfrac{2pi}{12}=cfrac{pi}{6}) (cfrac{5pi}{12}) (cfrac{8pi}{12}=cfrac{2pi}{3}) (cfrac{11pi}{12})
(2x+cfrac{pi}{6}) (0) (cfrac{pi}{2}) (pi) (cfrac{3pi}{2}) (2pi)
(sin(2x+frac{pi}{6})) (0) (1) (0) (-1) (0)
(f(x)) (1) (3) (1) (-1) (1)
(点的坐标) ((-cfrac{pi}{12},1)) ((cfrac{pi}{6},3)) ((cfrac{5pi}{12},1)) ((cfrac{2pi}{3},-1)) ((cfrac{11pi}{12},1))

效果图如下:

例1【题定周期的起止点】求函数(f(x)=2sin(2x+cfrac{pi}{6})+1)在区间([0,pi])上的函数图像;

分析:在上题作图的基础上修正如下即可,

(x) (0) (cfrac{2pi}{12}=cfrac{pi}{6}) (cfrac{5pi}{12}) (cfrac{8pi}{12}=cfrac{2pi}{3}) (cfrac{11pi}{12}) (pi)
(2x+cfrac{pi}{6}) (cfrac{pi}{6}) (cfrac{pi}{2}) (pi) (cfrac{3pi}{2}) (2pi) (cfrac{13pi}{6})
(sin(2x+frac{pi}{6})) (cfrac{1}{2}) (1) (0) (-1) (0) (cfrac{1}{2})
(f(x)) (2) (3) (1) (-1) (1) (2)
(坐标) ((0,2)) ((cfrac{pi}{6},3)) ((cfrac{5pi}{12},1)) ((cfrac{2pi}{3},-1)) ((cfrac{11pi}{12},1)) ((pi,2))

效果图如下:待整理;

原文地址:https://www.cnblogs.com/wanghai0666/p/10659481.html