11类RD Sharma解决方案–第六章三角函数图–练习6.3

绘制以下函数的图形：

问题1：y =罪² x

解决方案：

As we know that,

y = sin² x = $\frac{1- cos 2x}{2} = \frac{1}{2} - \frac{cos 2x}{2}$

$y - \frac{1}{2} = -\frac{cos 2x}{2}$

On shifting the origin at (0, 1/2), we get

X = x and Y = $y - \frac{1}{2}$

On substituting these values, we get

$Y = - \frac{cos 2X}{2}$

The maximum and minimum values of Y are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively and shift it by 1/2 to the up.

As the equation in the form of y = – f(x), the graph become inverted of y = f(x)

问题2：y = cos ² x

解决方案：

As we know that,

y = cos² x = $\frac{1+ cos 2x}{2} = \frac{1}{2} + \frac{cos 2x}{2}$

$y - \frac{1}{2} = \frac{cos 2x}{2}$

On shifting the origin at $(0, \frac{1}{2})$, we get

X = x and Y = $y - \frac{1}{2}$

On substituting these values, we get

$Y = \frac{cos 2X}{2}$

The maximum and minimum values of Y are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively and shift it by 1/2 to the up.

问题3：y =罪² $(x-\frac{\pi}{4})$

解决方案：

To obtain this graph y-0 = sin² $(x-\frac{\pi}{4})$

On shifting the origin at ($\frac{\pi}{4}$,0), we get

X = $x-\frac{\pi}{4}$ and Y = y – 0

On substituting these values, we get

Y = sin² X

First we draw the graph of Y = sin² X and shift it by π/4 to the right.

问题4：y =棕褐色2x

解决方案：

To obtain this graph y = tan 2x,

First we draw the graph of y = tan x and then divide the x-coordinates of the points where it crosses x-axis by 2.

问题5：y = 2棕褐色3x

解决方案：

To obtain this graph y = 2 tan 3x,

First we draw the graph of y = tan x and then divide the x-coordinates of the points where it crosses x-axis by 3.

Stretch the graph vertically by the factor of 2.

问题6：y = 2 cot 2x

解决方案：

To obtain this graph y = 2 cot 2x,

First we draw the graph of y = cot x and then divide the x-coordinates of the points where it crosses x-axis by 2.

Stretch the graph vertically by the factor of 2.

以相同比例绘制以下函数的图形：

问题7：y = cos ² x，y = cos $(2x-\frac{\pi}{3})$

解决方案：

Graph 1:

y = cos² x

As we know that,

y = cos² x = $\frac{1+ cos 2x}{2} = \frac{1}{2} + \frac{cos 2x}{2}$

$y - \frac{1}{2} = \frac{cos 2x}{2}$

On shifting the origin at (0, 1/2), we get

X = x and Y = $y - \frac{1}{2}$

On substituting these values, we get

$Y = \frac{cos 2X}{2}$

The maximum and minimum values of Y are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively and shift it by 1/2 to the up.

Graph 2:

To obtain this graph y-0 = cos (2x-$\frac{\pi}{3}$) = cos 2(x-$\frac{\pi}{6}$)

On shifting the origin at (π/6, 0), we get

X = $x-\frac{\pi}{6}$ and Y = y – 0

On substituting these values, we get

Y = cos 2X

First we draw the graph of Y = cos 2X and shift it by π/6 to the right.

The graph y = cos² x and y = cos $(2x-\frac{\pi}{3})$ are on same axes are as follows:

问题8：y = sin ² x，y = sin x

解决方案：

Graph 1:

y = sin² x

As we know that,

y = sin² x = $\frac{1- cos 2x}{2} = \frac{1}{2} - \frac{cos 2x}{2}$

$y - \frac{1}{2} = -\frac{cos 2x}{2}$

On shifting the origin at (0, $\frac{1}{2}$), we get

X = x and Y = $y - \frac{1}{2}$

On substituting these values, we get

$Y = - \frac{cos 2X}{2}$

The maximum and minimum values of Y are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively and shift it by 1/2 to the up.

As the equation in the form of y = – f(x), the graph become inverted of y = f(x)

Graph 2:

y = sin x

The graph y = sin² x and y = sin x are on same axes are as follows:

问题9：y =棕褐色x，y =棕褐色² x

解决方案：

Graph 1:

y = tan x

Graph 2:

y = tan² x

The graph y = tan x and y = tan² x are on same axes are as follows:

问题10：y = tan 2x，y = tan x

解决方案：

Graph 1:

To obtain this graph y = tan 2x,

First we draw the graph of y = tan x and then divide the x-coordinates of the points where it crosses x-axis by 2.

Graph 2:

y = tan x

The graph y = tan 2x and y = tan x are on same axes are as follows: