Rd Sharma Xi 2018 Solutions for Class 12 Science Math Chapter 6 Graphs Of Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Graphs Of Trigonometric Functions are extremely popular among Class 12 Science students for Math Graphs Of Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 12 Science Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 6.13:

Question 1:

Sketch the graphs of the following functions:
f(x) = 2 cosec πx

Answer:

f(x) = 2 cosec πx

 

Page No 6.13:

Question 2:

Sketch the graphs of the following functions:
f
(x) = 3 sec x

Answer:

f(x) = 3 sec x

 

Page No 6.13:

Question 3:

Sketch the graphs of the following functions:
f
(x) = cot 2x

Answer:

f(x) = cot 2x

 

Page No 6.13:

Question 4:

Sketch the graphs of the following functions:
f
(x) = 2 sec πx

Answer:

f(x) = 2 sec πx

 

Page No 6.13:

Question 5:

Sketch the graphs of the following functions:
f
(x) = tan2 x

Answer:

f(x) = tan2 x

 

Page No 6.13:

Question 6:

Sketch the graphs of the following functions:
f
(x) = cot2 x

Answer:

f(x) = cot2 x

 

Page No 6.13:

Question 7:

Sketch the graphs of the following functions:
fx=cotπx2

Answer:

fx=cotπx2

 

Page No 6.13:

Question 8:

Sketch the graphs of the following functions:
f
(x) = sec2 x

Answer:

f(x) = sec2 x

 

Page No 6.13:

Question 9:

Sketch the graphs of the following functions:
f(x) = cosec2 x

Answer:

f(x) = cosec2 x

 

Page No 6.13:

Question 10:

Sketch the graphs of the following functions:
f
(x) = tan 2x

Answer:

Step I- We find the value of c and a by comparing y = 2 tan 2x with y = c tan ax, i.e. c = 1 and a = 2.
Step II- Then, we draw the graph of y =  tan x and mark the point where it crosses the x-axis.
Step III- Divide the x-coordinates of the points where y = tan crosses x-axis by 2(i.e. a = 2) and mark the maximum value (i.e. c = 1) and minimum value (i.e.-c = -1).
Then , we obtain the following graph:



Page No 6.5:

Question 1:

Sketch the graphs of the following functions:

(i) f(x) = 2 sin x, 0 ≤ x ≤ π
(ii) gx=3 sin x-π4, 0x5π4
(iii) hx=2 sin 3x, 0x2π3
(iv) ϕx=2 sin 2x-π3, 0x7π5
(v) ψx=4 sin 3x-π4, 0x2π
(vi) θx=sin x2-π4, 0x4π
(vii) ux=sin2 x, 0x2π vx=sin x, 0x2π
(viii) f(x) = 2 sin πx, 0 ≤ x ≤ 2

Answer:

The graphs of the following functions are:

(i) f(x) = 2 sin x, 0 ≤ x ≤ π
 

x 0 π
f(x) = 2 sin x 0 0

 

(ii) gx=3 sin x-π4, 0x5π4
 
x π4 5π4
gx=3 sin x-π4 0 0

 

(iii) hx=2 sin 3x, 0x2π3
 
x 0 π3 2π3
hx=2 sin 3x 0 0 0

 

(iv) ϕx=2 sin 2x-π3, 0x7π5
 
x π6 4π6
ϕx=2 sin 2x-π3 0 0

 

(v) ψx=4 sin 3x-π4, 0x2π
 
x π4 7π12
ψx=4 sin 3x-π4 0 0

 

(vi) θx=sin x2-π4, 0x4π
 
x π2 5π2
θx=sin x2-π4 0 0

 

(vii) ux=sin2 x, 0x2πvx=sin x, 0x2π
 
x 0 π
ux=sin2 x 0 0

 
 
x 0 π
vx=sin x 0 0

 

(viii) f(x) = 2 sin πx, 0 ≤ x ≤ 2
 
x 0 1
f(x) = 2 sin πx 0 0

 

Page No 6.5:

Question 2:

Sketch the graphs of the following pairs of functions on the same axes:
(i) fx=sin x, gx=sin x+π4
(ii) f(x) = sin x, g(x) = sin 2x
(iii) f(x) = sin 2x, g(x) = 2 sin x
(iv) fx=sinx2, gx=sin x

Answer:

(i) fx=sin x, gx=sin x+π4
Clearly, sin x and sin x+π4 is a periodic function with period 2π.

The graphs of fx=sin x and gx=sin x+π4 on different axes are shown below:

 

 

If these two graphs are drawn on the same axes, then the graph is shown below.

 

(ii) f(x) = sin x, g(x) = sin 2x

Clearly, sin x and sin 2x is a periodic function with period 2π and π, respectively.

The graphs of f(x) = sin x and g(x) = sin 2x on different axes are shown below:

 

 

If these two graphs are drawn on the same axes, then the graph is shown below.

 

(iii) f(x) = sin 2x, g(x) = 2 sin x

Clearly, sin 2x and 2 sin x is a periodic function with period π and 2π, respectively.

The graphs of f(x) = sin 2x and g(x) = 2 sin x on different axes are shown below:

 

 

If these two graphs are drawn on the same axes, then the graph is shown below.

 

(iv) fx=sinx2, gx=sin x

Clearly, sin x2 and sin x is a periodic function with period 4π and 2π, respectively.

The graphs of f(x) = sin x2 and g(x) = sin x on different axes are shown below:

 

 

If these two graphs are drawn on the same axes, then the graph is shown below.

 



Page No 6.8:

Question 1:

Sketch the graphs of the following trigonometric functions:
(i) fx=cosx-π4
(ii) gx=cosx+π4
(iii) h(x) = cos2 2x
(iv) ϕx=2 cosx-π6
(v) ψ(x) = cos 3x
(vi) ux=cos2x2
(vii) f(x) = cos π x
(viii) g(x) = cos 2π x

Answer:

(i)
y=cosx-π4 y-0=cosx-π4                    ...(i)On shifting the origin at π4,0, we get:x=X+π4 and y=Y+0On subsitituting the values in (i)  we get:Y=cosXThen, we draw the graph of Y=cosX and shift it by π4 to the right.
Then, we obtain the following graph:


 (ii)
y=cosx+π4 y-0=cosx+π4                   ...(i)On shifting the origin at -π4,0, we get:x=X-π4 and y=Y+0On subsitituting the values in (i), we get:Y=cosXThen, we draw the graph of Y=cosX and shift it by π4 to the left.
Then, we obtain the following graph:


(iii)
y=cos2 2x

The following graph is:

 

(iv)
y=2 cosx-π6 y-0=2 cosx-π6                   ...(i)On shifting the origin at π6,0, we get:x=X+π6 and y=Y+0On subsitituting the values in (i), we get:Y=2 cos XThen, we draw the graph of Y=cosX and shift it by π6 to the right.
Then, we obtain the following graph:
 
 

(v)
y=cos 3x

The following graph is:
 
 

(vi)
y=cos2 x2

The following graph is:
 
 

(vii)
y=cos πx

The following graph is:
 
 

(viii)
y=cos 2πx

The following graph is:
 
 

Page No 6.8:

Question 2:

Sketch the graphs of the following curves on the same scale and the same axes:
(i) y=cos x and y=cos x-π4
(ii) y=cos 2x and y=cos 2x-π4
(iii) y=cos x and y=cosx2
(iv) y=cos2x and y=cos x 

Answer:

(i)
First, we draw the graph of y = cos x.
Let us now draw the graph of y=cosx-π4.
y=cosx-π4 y-0=cosx-π4            ...(i)On shifting the origin at π4,0, we get:x=X+π4 and y=Y+0On subsitituting the values in (i), we get:Y=cosXThen, we draw the graph of Y=cosX and shift it by π4 to the right.
 Then, we will obtain the following graph:



(ii)
First, we draw the graph of y = cos 2x.
Let us now draw the graph of y=cos2x-π4.
y=cos2x-π4 y-0=cos2x-π4             ...(i)On shifting the origin at π4,0, we get:x=X+π4 and y=Y+0On subsitituting the values in (i), we get:Y=cos2XThen, we draw the graph of Y=cos2X and shift it by π4 to the right.
 Then, we will obtain the following graph:



(iii)

First, we draw the graph of y = cos x.
Let us now draw the graph of y=cosx2.
   y=cos12x
 Then, we will obtain the following graph:


(iv)
First, we draw the graph of y = cos2 x.
Let us now draw the graph of y = cos x.
 
Then, we will obtain the following graph:

 



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