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Class 11 RD Sharma Solutions – Chapter 6 Graphs of Trigonometric Functions – Exercise 6.3

Last Updated : 08 May, 2021
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Sketch the graphs of the following functions:

Question 1: y = sin2 x

Solution:

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)

Question 2: y = cos2 x

Solution:

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.

Question 3: y = sin2 (x-\frac{\pi}{4})

Solution:

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.

Question 4: y = tan 2x

Solution:

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.

Question 5: y = 2 tan 3x

Solution:

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.

Question 6: y = 2 cot 2x

Solution:

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.

Sketch the graphs of the following functions on the same scale:

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

Solution:

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:

Question 8: y = sin2 x, y = sin x

Solution:

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:

Question 9: y = tan x, y = tan 2x

Solution:

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:

Question 10: y = tan 2x, y = tan x

Solution:

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:



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