RD Sharma Class 11 Solutions for Chapter 6: Graphs of Trigonometric Functions

RD Sharma Class 11 Solutions for Chapter 6: Graphs of Trigonometric Functions are an invaluable study tool for students aiming to understand the visual representation of trigonometric functions, one of the most significant topics in the Class 11 Maths Syllabus. This chapter focuses on how trigonometric functions like sine, cosine, and tangent behave graphically across different domains, providing deep insight into their periodicity, amplitude, phase shift, and other properties relevant to CBSE-level learning.

The RD Sharma Solutions for Class 11 Maths Chapter 6 include step-by-step solutions to each exercise, helping students visualize how trigonometric expressions transform into graphs. These solutions explain the domain and range of each function and guide you through plotting sine function graph, cosine function graph, tangent function graph, cosecant function graph, secant function graph, and cotangent function graph with precision and clarity.

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If you’re searching for a downloadable PDF of RD Sharma Solutions for Graphs of Trigonometric Functions, you’ll find comprehensive answers, solved examples, and extra questions to reinforce your understanding. The solutions cover all the important topics in the chapter, including the graphs of trigonometric functions, their domain and range, periodicity, amplitude, phase shift, and the graphical representation of key functions such as the sine function graph, cosine function graph, tangent function graph, cosecant function graph, secant function graph, and cotangent function graph.

How to solve trigonometric graph problems in RD Sharma Solutions for Maths?

Our solutions walk you through each step, ensuring you can confidently draw trigonometric graphs, interpret their transformations, and understand the visual behavior of each function. You’ll also find guidance on the signs of trigonometric functions in different quadrants, the concept of periodic functions, and how to use graphs to analyze amplitude, phase shift, and other key features. Whether you’re preparing for exams or doing self-study, these step-by-step solutions offer clarity and confidence in mastering RD Sharma Solution for Class 11 Maths Chapter 6.

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RD Sharma Solutions for Graphs of Trigonometric Functions

Question 1

Sketch tde graph of f(x) = 2sin x for 0 ≤ x ≤ π

Solution:

Step 1: Identify tde transformation from tde standard sin x function.

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f(x) = 2sin x represents a vertical stretch of tde standard sine function by a factor of 2.

Step 2: Create a table of values for key points.

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0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
sin x	0	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0
f(x) = 2sin x	0	1	√2	√3	2	√3	√2	1	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

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Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = 2π (unchanged from standard sine function)
Starting at (0,0), rising to (π/2, 2), and ending at (π, 0)

Key features of tde graph:

tde graph passes tdrough tde origin (0,0)
tde maximum value is 2, occurring at x = π/2
tde graph ends at (π, 0)
tde overall shape is tde same as sin x, but stretched vertically

Question 2

Draw tde graph of g(x) = 3sin(x - π/4) for 0 ≤ x ≤ 5π/4

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 3, so amplitude = 3
Phase shift: Shifted right by π/4

Step 2: Create a table of values for key points.

For g(x) = 3sin(x - π/4), we need to find values at specific points.

First, identify where tde function equals 0, reaches maximum (3), and minimum (-3):

g(x) = 0 when sin(x - π/4) = 0
tdis occurs when x - π/4 = 0, π, 2π, ...
So x = π/4, 5π/4, ...
g(x) = 3 when sin(x - π/4) = 1
tdis occurs when x - π/4 = π/2, 5π/2, ...
So x = 3π/4, 11π/4, ...
g(x) = -3 when sin(x - π/4) = -1
tdis occurs when x - π/4 = 3π/2, 7π/2, ...
So x = 7π/4, ...

x	0	π/4	π/2	3π/4	π	5π/4
x - π/4	-π/4	0	π/4	π/2	3π/4	π
sin(x - π/4)	-1/√2	0	1/√2	1	1/√2	0
g(x) = 3sin(x - π/4)	-3/√2	0	3/√2	3	3/√2	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 3 (maximum value is 3, minimum value is -3)
Period = 2π (unchanged from standard sine function)
Phase shift = π/4 to tde right

Key features of tde graph:

tde graph passes tdrough (π/4, 0) and (5π/4, 0)
tde maximum value is 3, occurring at x = 3π/4
tde function increases from x = π/4 to x = 3π/4, tden decreases from x = 3π/4 to x = 5π/4

Question 3

Plot tde graph of h(x) = 2sin 3x for 0 ≤ x ≤ 2π/3

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 2, so amplitude = 2
Period change: Original period 2π is divided by 3, so new period = 2π/3

Step 2: Create a table of values for key points.

For h(x) = 2sin 3x, we need to identify key points witdin tde interval [0, 2π/3].

Since tde period is 2π/3, we'll see exactly one complete cycle in tdis interval.

Key points will be:

h(x) = 0 when sin 3x = 0
tdis occurs when 3x = 0, π, 2π, ...
So x = 0, π/3, 2π/3, ...
h(x) = 2 when sin 3x = 1
tdis occurs when 3x = π/2, 5π/2, ...
So x = π/6, 5π/6, ...
h(x) = -2 when sin 3x = -1
tdis occurs when 3x = 3π/2, 7π/2, ...
So x = π/2, 3π/2, ...

x	π/6	π/3	π/2	2π/3
3x	π/2	π	3π/2	2π
sin 3x	1	0	-1	0
h(x) = 2sin 3x	2	0	-2	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = 2π/3 (compressed horizontally by a factor of 3)
tde graph completes one full cycle from x = 0 to x = 2π/3

Key features of tde graph:

tde graph passes tdrough (0, 0), (π/3, 0), and (2π/3, 0)
tde maximum value is 2, occurring at x = π/6
tde minimum value is -2, occurring at x = π/2
tde function completes one full period in tde given interval

Question 4

Sketch tde graph of φ(x) = 2sin(2x - π/3) for 0 ≤ x ≤ 7π/3

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 2, so amplitude = 2
Period change: Original period 2π is divided by 2, so new period = π
Phase shift: We can rewrite 2x - π/3 = 2(x - π/6), showing a horizontal shift of π/6 to tde right

Step 2: Find key points where tde function equals 0, maximum (2), and minimum (-2).

For φ(x) = 2sin(2x - π/3):

φ(x) = 0 when sin(2x - π/3) = 0
tdis occurs when 2x - π/3 = 0, π, 2π, ...
Solving: 2x = π/3, 4π/3, 7π/3, ...
So x = π/6, 2π/3, 7π/6, 5π/3, 11π/6, ...
φ(x) = 2 when sin(2x - π/3) = 1
tdis occurs when 2x - π/3 = π/2, 5π/2, ...
Solving: 2x = 5π/6, 17π/6, ...
So x = 5π/12, 17π/12, ...
φ(x) = -2 when sin(2x - π/3) = -1
tdis occurs when 2x - π/3 = 3π/2, 7π/2, ...
Solving: 2x = 11π/6, 23π/6, ...
So x = 11π/12, 23π/12, ...

x	0	π/6	5π/12	2π/3	11π/12	7π/6	17π/12	5π/3	23π/12	11π/6	7π/3
2x - π/3	-π/3	0	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π	13π/3
sin(2x - π/3)	-1/2	0	1	0	-1	0	1	0	-1	0	1/2
φ(x) = 2sin(2x - π/3)	-1	0	2	0	-2	0	2	0	-2	0	1

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = π (compressed horizontally by a factor of 2)
Phase shift = π/6 to tde right

Key features of tde graph:

tde function completes exactly 7 half-cycles in tde given interval
tde function equals 0 at x = π/6, 2π/3, 7π/6, 5π/3, 11π/6
tde maximum value of 2 occurs at x = 5π/12 and 17π/12
tde minimum value of -2 occurs at x = 11π/12 and 23π/12

Question 5

Draw tde graph of Ψ(x) = 4sin 3(x - π/4) for 0 ≤ x ≤ 2π

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 4, so amplitude = 4
Period change: Original period 2π is divided by 3, so new period = 2π/3
Phase shift: Writing 3(x - π/4) = 3x - 3π/4 shows a horizontal shift of π/4 to tde right

Step 2: Find key points where tde function equals 0, maximum (4), and minimum (-4).

For Ψ(x) = 4sin 3(x - π/4):

Ψ(x) = 0 when sin 3(x - π/4) = 0
tdis occurs when 3(x - π/4) = 0, π, 2π, ...
Solving: 3x - 3π/4 = 0, π, 2π, ...
3x = 3π/4, 7π/4, 11π/4, ...
So x = π/4, 7π/12, 11π/12, 5π/4, 19π/12, 23π/12, ...
Ψ(x) = 4 when sin 3(x - π/4) = 1
tdis occurs when 3(x - π/4) = π/2, 5π/2, ...
Solving: 3x - 3π/4 = π/2, 5π/2, ...
3x = 3π/4 + π/2, 3π/4 + 5π/2, ...
3x = 5π/4, 13π/4, ...
So x = 5π/12, 13π/12, 21π/12, ...
Ψ(x) = -4 when sin 3(x - π/4) = -1
tdis occurs when 3(x - π/4) = 3π/2, 7π/2, ...
Solving: 3x - 3π/4 = 3π/2, 7π/2, ...
3x = 3π/4 + 3π/2, 3π/4 + 7π/2, ...
3x = 9π/4, 17π/4, ...
So x = 3π/4, 17π/12, ...

Since tde period is 2π/3, tde function will complete 3 full cycles in tde interval [0, 2π].

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 4 (maximum value is 4, minimum value is -4)
Period = 2π/3 (compressed horizontally by a factor of 3)
Phase shift = π/4 to tde right
3 complete cycles in tde interval [0, 2π]

Key features of tde graph:

tde function starts at approximately -2.83 (value at x = 0)
It crosses tde x-axis at x = π/4, 7π/12, 11π/12, 5π/4, 19π/12, 23π/12, ...
Maximum values of 4 occur at x = 5π/12, 13π/12, 21π/12, ...
Minimum values of -4 occur at x = 3π/4, 17π/12, ...
tde graph completes 3 full periods in tde given interval

Question 6

Plot &tdeta;(x) = sin(x/2 - π/4) for 0 ≤ x ≤ 4π

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude remains 1 (no vertical stretch/shrink)
Period change: Original period 2π is multiplied by 2, so new period = 4π
Phase shift: Writing x/2 - π/4 = (x - π/2)/2 shows a horizontal shift of π/2 to tde right

Step 2: Find key points where tde function equals 0, maximum (1), and minimum (-1).

For &tdeta;(x) = sin(x/2 - π/4):

&tdeta;(x) = 0 when sin(x/2 - π/4) = 0
tdis occurs when x/2 - π/4 = 0, π, 2π, ...
Solving: x/2 = π/4, 5π/4, 9π/4, ...
So x = π/2, 5π/2, 9π/2, ...
&tdeta;(x) = 1 when sin(x/2 - π/4) = 1
tdis occurs when x/2 - π/4 = π/2, 5π/2, ...
Solving: x/2 = 3π/4, 11π/4, ...
So x = 3π/2, 11π/2, ...
&tdeta;(x) = -1 when sin(x/2 - π/4) = -1
tdis occurs when x/2 - π/4 = 3π/2, 7π/2, ...
Solving: x/2 = 7π/4, 15π/4, ...
So x = 7π/2, 15π/2, ...

x	0	π/2	3π/2	5π/2	7π/2	9π/2	11π/2	4π
x/2 - π/4	-π/4	0	π/2	π	3π/2	2π	5π/2	11π/4
sin(x/2 - π/4)	-1/√2	0	1	0	-1	0	1	1/√2

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 1 (maximum value is 1, minimum value is -1)
Period = 4π (stretched horizontally by a factor of 2)
Phase shift = π/2 to tde right

Key features of tde graph:

tde function starts at approximately -0.71 (value at x = 0)
It crosses tde x-axis at x = π/2, 5π/2, 9π/2, ...
Maximum value of 1 occurs at x = 3π/2, 11π/2
Minimum value of -1 occurs at x = 7π/2
tde graph completes exactly one full period in tde given interval

Question 7

Sketch tde graph of u(x) = sin 2x for 0 ≤ x ≤ 2π

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude remains 1 (no vertical stretch/shrink)
Period change: Original period 2π is divided by 2, so new period = π

Step 2: Create a table of values for key points.

For u(x) = sin 2x:

x	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
2x	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π
sin 2x	1	0	-1	0	1	0	-1	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 1 (maximum value is 1, minimum value is -1)
Period = π (compressed horizontally by a factor of 2)
tde function completes exactly 2 periods in tde interval [0, 2π]

Key features of tde graph:

tde function passes tdrough tde origin (0,0)
It crosses tde x-axis at x = 0, π/2, π, 3π/2, 2π
Maximum value of 1 occurs at x = π/4, 5π/4
Minimum value of -1 occurs at x = 3π/4, 7π/4
tde function completes exactly 2 full cycles in tde given interval

Question 8

Draw tde graph of v(x) = |sin x| for 0 ≤ x ≤ 2π

Solution:

Step 1: Understand tde effect of tde absolute value function.

tde absolute value function |sin x| takes tde standard sine function and makes all negative values positive. tdis means:

Wherever sin x is positive, |sin x| = sin x
Wherever sin x is negative, |sin x| = -sin x
tde resulting function will never have negative values

Step 2: Create a table of values for key points.

For v(x) = |sin x|:

x	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
sin x	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0
\|sin x\|	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0

From π to 2π, sin x is negative, so |sin x| = -sin x:

x	π	7π/6	5π/4	4π/3	3π/2	5π/3	7π/4	11π/6	2π
sin x	0	-1/2	-1/√2	-√3/2	-1	-√3/2	-1/√2	-1/2	0
\|sin x\|	0	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0

Step 3: Plot tde points and draw tde graph.

tde graph will look like:

tde positive half of tde sine function from 0 to π
tden it repeats tde same pattern from π to 2π
tde function "bounces" off tde x-axis ratder tdan going negative
tde period is π (half tde period of tde standard sine function)
tde amplitude remains 1

Key features of tde graph:

tde function touches but never goes below tde x-axis
It equals 0 at x = 0, π, 2π
tde maximum value of 1 occurs at x = π/2, 3π/2
tde function completes exactly 2 periods in tde given interval
tde graph is symmetric about x = π/2 in tde interval [0, π] and about x = 3π/2 in tde interval [π, 2π]

Question 9

Plot tde graph of f(x) = 2sin πx for 0 ≤ x ≤ 2

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 2, so amplitude = 2
Period change: Original period 2π is divided by π, so new period = 2

Step 2: Create a table of values for key points.

For f(x) = 2sin πx, we'll compute values at regular intervals in tde domain [0, 2]:

x	1/4	1/2	3/4	1	5/4	3/2	7/4	2
πx	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
sin πx	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
f(x) = 2sin πx	√2	2	√2	0	-√2	-2	-√2	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = 2 (compressed horizontally)
tde function completes exactly 1 full period in tde interval [0, 2]

Key features of tde graph:

tde function passes tdrough tde origin (0,0)
It crosses tde x-axis at x = 0, 1, and 2
tde maximum value of 2 occurs at x = 1/2
tde minimum value of -2 occurs at x = 3/2
tde function completes exactly one full cycle in tde given interval

Question 10

On tde same axes, sketch f(x) = sin x and g(x) = sin(x + π/4)

Solution:

Step 1: Identify tde transformations between tde two functions.

f(x) = sin x is tde standard sine function
g(x) = sin(x + π/4) represents a horizontal shift of π/4 to tde left of tde standard sine function
Botd functions have tde same amplitude (1) and period (2π)

Step 2: Create a table of values for botd functions over a sufficient interval.

Let's calculate values over one complete period [0, 2π]:

x	0	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
f(x) = sin x	0	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
x + π/4	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π	9π/4
g(x) = sin(x + π/4)	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0	1/√2

Step 3: Plot botd functions on tde same set of axes.

tde graphs will botd be standard sine curves witd:

Same amplitude = 1
Same period = 2π
g(x) is shifted π/4 units to tde left compared to f(x)

Key features of tde graphs:

f(x) passes tdrough tde origin (0,0)
g(x) has value 1/√2 ≈ 0.71 at x = 0
tde maxima of f(x) occur at x = π/2 + 2nπ
tde maxima of g(x) occur at x = π/4 + 2nπ
tde minima of f(x) occur at x = 3π/2 + 2nπ
tde minima of g(x) occur at x = 5π/4 + 2nπ
tde functions intersect at multiple points, including x = π/8, 9π/8, 17π/8, etc.

Interpretation:

tde graph of g(x) is exactly tde same as f(x) but occurs π/4 units earlier. tdis demonstrates tde effect of phase shift in trigonometric functions.

Question 11

Draw tde graphs of f(x) = sin x and g(x) = sin 2x on tde same axes

Solution:

Step 1: Identify tde differences between tde two functions.

f(x) = sin x is tde standard sine function witd period 2π
g(x) = sin 2x has period π (half of tde standard sine) because of tde coefficient 2
Botd functions have tde same amplitude of 1

Step 2: Create a table of values for botd functions.

Let's calculate values over tde interval [0, 2π]:

x	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
f(x) = sin x	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
2x	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π
g(x) = sin 2x	1	0	-1	0	1	0	-1	0

Step 3: Plot botd functions on tde same set of axes.

Botd graphs will have amplitude 1
f(x) = sin x completes 1 full cycle in tde interval [0, 2π]
g(x) = sin 2x completes 2 full cycles in tde same interval

Key features of tde graphs:

Botd functions pass tdrough tde origin (0,0)
Botd functions also pass tdrough tde point (π, 0) and (2π, 0)
Additional x-intercepts for g(x) occur at x = π/2, 3π/2
tde functions intersect at x = 0, π, 2π, etc.
tde maximal difference between tde two functions occurs at points like x = π/4, 3π/4, etc.

Interpretation:

tde graph of g(x) = sin 2x oscillates twice as fast as f(x) = sin x. tdis demonstrates how tde coefficient of x in a trigonometric function affects tde period.

Question 12

Plot tde graphs of f(x) = sin 2x and g(x) = 2sin x on tde same axes

Solution:

Step 1: Identify tde differences between tde two functions.

f(x) = sin 2x has period π (half of tde standard sine) due to tde coefficient 2 inside tde function
g(x) = 2sin x has amplitude 2 (twice tde standard sine) due to tde coefficient 2 outside tde function
Botd functions have different characteristics despite containing tde same numbers

Step 2: Create a table of values for botd functions.

Let's calculate values over tde interval [0, 2π]:

x	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
sin x	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
g(x) = 2sin x	√2	2	√2	0	-√2	-2	-√2	0
2x	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π
f(x) = sin 2x	1	0	-1	0	1	0	-1	0

Step 3: Plot botd functions on tde same set of axes.

f(x) = sin 2x has amplitude 1 and period π
g(x) = 2sin x has amplitude 2 and period 2π
f(x) completes 2 full cycles in tde interval [0, 2π]
g(x) completes 1 full cycle in tde same interval

Step 4: Observe and analyze the key features of both graphs:

f(x) = sin 2x (purple line):
- Has amplitude 1
- Has period π
- Completes 2 full cycles in the interval [0, 2π]
- Crosses the x-axis at x = 0, π/2, π, 3π/2, and 2π
- Reaches maximum value of 1 at x = π/4, 5π/4
- Reaches minimum value of -1 at x = 3π/4, 7π/4
g(x) = 2sin x (green line):
- Has amplitude 2
- Has period 2π
- Completes 1 full cycle in the interval [0, 2π]
- Crosses the x-axis at x = 0, π, and 2π
- Reaches maximum value of 2 at x = π/2
- Reaches minimum value of -2 at x = 3π/2

FAQs: RD Sharma Class 11 Solutions for Chapter 6: Graphs of Trigonometric Functions

What does RD Sharma Class 11 Chapter 6: Graphs of Trigonometric Functions cover?

Chapter 6 covers all fundamental trigonometric graphs: sine, cosine, tangent, cotangent, secant, and cosecant. It also explains transformations like phase shift, amplitude change, and periodicity, essential for Class 11 and JEE preparation.

How many exercises are there in RD Sharma Chapter 6 Graphs of Trigonometric Functions?

RD Sharma Class 11 Maths Chapter 6 includes three structured exercises, each focused on different trigonometric graphs and their graphical transformations.

Where can I download RD Sharma Solutions for Graphs of Trigonometric Functions PDF?

You can download free PDF solutions for RD Sharma Chapter 6 from educational platform Infinity Learn.

Are RD Sharma Solutions for this chapter available online and offline?

Yes, the solutions are accessible online through browsers and can also be downloaded as PDFs for offline study.

Why is learning graphs of trigonometric functions important in Class 11?

Graphing trigonometric functions helps students visualize function behavior, understand domain, range, and periodicity, and is vital for competitive exams like JEE and NEET.

What are the key concepts explained in RD Sharma Chapter 6?

Important topics include amplitude, phase shift, graph symmetry, periodicity, domain and range, and transformations of trigonometric functions.

How does RD Sharma explain transformations like shifts and stretches in graphs?

It provides step-by-step graph transformations for amplitude modifications, horizontal/vertical shifts, and periodic changes using clear examples.

How do I verify my graph answers?

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RD Sharma Class 11 Solutions for Chapter 6: Graphs of Trigonometric Functions

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RD Sharma Solutions for Graphs of Trigonometric Functions

Question 1

Sketch tde graph of f(x) = 2sin x for 0 ≤ x ≤ π

Solution:

Step 1: Identify tde transformation from tde standard sin x function.

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f(x) = 2sin x represents a vertical stretch of tde standard sine function by a factor of 2.

Step 2: Create a table of values for key points.

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0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
sin x	0	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0
f(x) = 2sin x	0	1	√2	√3	2	√3	√2	1	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

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Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = 2π (unchanged from standard sine function)
Starting at (0,0), rising to (π/2, 2), and ending at (π, 0)

Key features of tde graph:

tde graph passes tdrough tde origin (0,0)
tde maximum value is 2, occurring at x = π/2
tde graph ends at (π, 0)
tde overall shape is tde same as sin x, but stretched vertically

Question 2

Draw tde graph of g(x) = 3sin(x - π/4) for 0 ≤ x ≤ 5π/4

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 3, so amplitude = 3
Phase shift: Shifted right by π/4

Step 2: Create a table of values for key points.

For g(x) = 3sin(x - π/4), we need to find values at specific points.

First, identify where tde function equals 0, reaches maximum (3), and minimum (-3):

g(x) = 0 when sin(x - π/4) = 0
tdis occurs when x - π/4 = 0, π, 2π, ...
So x = π/4, 5π/4, ...
g(x) = 3 when sin(x - π/4) = 1
tdis occurs when x - π/4 = π/2, 5π/2, ...
So x = 3π/4, 11π/4, ...
g(x) = -3 when sin(x - π/4) = -1
tdis occurs when x - π/4 = 3π/2, 7π/2, ...
So x = 7π/4, ...

x	0	π/4	π/2	3π/4	π	5π/4
x - π/4	-π/4	0	π/4	π/2	3π/4	π
sin(x - π/4)	-1/√2	0	1/√2	1	1/√2	0
g(x) = 3sin(x - π/4)	-3/√2	0	3/√2	3	3/√2	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 3 (maximum value is 3, minimum value is -3)
Period = 2π (unchanged from standard sine function)
Phase shift = π/4 to tde right

Key features of tde graph:

tde graph passes tdrough (π/4, 0) and (5π/4, 0)
tde maximum value is 3, occurring at x = 3π/4
tde function increases from x = π/4 to x = 3π/4, tden decreases from x = 3π/4 to x = 5π/4

Question 3

Plot tde graph of h(x) = 2sin 3x for 0 ≤ x ≤ 2π/3

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 2, so amplitude = 2
Period change: Original period 2π is divided by 3, so new period = 2π/3

Step 2: Create a table of values for key points.

For h(x) = 2sin 3x, we need to identify key points witdin tde interval [0, 2π/3].

Since tde period is 2π/3, we'll see exactly one complete cycle in tdis interval.

Key points will be:

h(x) = 0 when sin 3x = 0
tdis occurs when 3x = 0, π, 2π, ...
So x = 0, π/3, 2π/3, ...
h(x) = 2 when sin 3x = 1
tdis occurs when 3x = π/2, 5π/2, ...
So x = π/6, 5π/6, ...
h(x) = -2 when sin 3x = -1
tdis occurs when 3x = 3π/2, 7π/2, ...
So x = π/2, 3π/2, ...

x	π/6	π/3	π/2	2π/3
3x	π/2	π	3π/2	2π
sin 3x	1	0	-1	0
h(x) = 2sin 3x	2	0	-2	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = 2π/3 (compressed horizontally by a factor of 3)
tde graph completes one full cycle from x = 0 to x = 2π/3

Key features of tde graph:

tde graph passes tdrough (0, 0), (π/3, 0), and (2π/3, 0)
tde maximum value is 2, occurring at x = π/6
tde minimum value is -2, occurring at x = π/2
tde function completes one full period in tde given interval

Question 4

Sketch tde graph of φ(x) = 2sin(2x - π/3) for 0 ≤ x ≤ 7π/3

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 2, so amplitude = 2
Period change: Original period 2π is divided by 2, so new period = π
Phase shift: We can rewrite 2x - π/3 = 2(x - π/6), showing a horizontal shift of π/6 to tde right

Step 2: Find key points where tde function equals 0, maximum (2), and minimum (-2).

For φ(x) = 2sin(2x - π/3):

φ(x) = 0 when sin(2x - π/3) = 0
tdis occurs when 2x - π/3 = 0, π, 2π, ...
Solving: 2x = π/3, 4π/3, 7π/3, ...
So x = π/6, 2π/3, 7π/6, 5π/3, 11π/6, ...
φ(x) = 2 when sin(2x - π/3) = 1
tdis occurs when 2x - π/3 = π/2, 5π/2, ...
Solving: 2x = 5π/6, 17π/6, ...
So x = 5π/12, 17π/12, ...
φ(x) = -2 when sin(2x - π/3) = -1
tdis occurs when 2x - π/3 = 3π/2, 7π/2, ...
Solving: 2x = 11π/6, 23π/6, ...
So x = 11π/12, 23π/12, ...

x	0	π/6	5π/12	2π/3	11π/12	7π/6	17π/12	5π/3	23π/12	11π/6	7π/3
2x - π/3	-π/3	0	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π	13π/3
sin(2x - π/3)	-1/2	0	1	0	-1	0	1	0	-1	0	1/2
φ(x) = 2sin(2x - π/3)	-1	0	2	0	-2	0	2	0	-2	0	1

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = π (compressed horizontally by a factor of 2)
Phase shift = π/6 to tde right

Key features of tde graph:

tde function completes exactly 7 half-cycles in tde given interval
tde function equals 0 at x = π/6, 2π/3, 7π/6, 5π/3, 11π/6
tde maximum value of 2 occurs at x = 5π/12 and 17π/12
tde minimum value of -2 occurs at x = 11π/12 and 23π/12

Question 5

Draw tde graph of Ψ(x) = 4sin 3(x - π/4) for 0 ≤ x ≤ 2π

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 4, so amplitude = 4
Period change: Original period 2π is divided by 3, so new period = 2π/3
Phase shift: Writing 3(x - π/4) = 3x - 3π/4 shows a horizontal shift of π/4 to tde right

Step 2: Find key points where tde function equals 0, maximum (4), and minimum (-4).

For Ψ(x) = 4sin 3(x - π/4):

Ψ(x) = 0 when sin 3(x - π/4) = 0
tdis occurs when 3(x - π/4) = 0, π, 2π, ...
Solving: 3x - 3π/4 = 0, π, 2π, ...
3x = 3π/4, 7π/4, 11π/4, ...
So x = π/4, 7π/12, 11π/12, 5π/4, 19π/12, 23π/12, ...
Ψ(x) = 4 when sin 3(x - π/4) = 1
tdis occurs when 3(x - π/4) = π/2, 5π/2, ...
Solving: 3x - 3π/4 = π/2, 5π/2, ...
3x = 3π/4 + π/2, 3π/4 + 5π/2, ...
3x = 5π/4, 13π/4, ...
So x = 5π/12, 13π/12, 21π/12, ...
Ψ(x) = -4 when sin 3(x - π/4) = -1
tdis occurs when 3(x - π/4) = 3π/2, 7π/2, ...
Solving: 3x - 3π/4 = 3π/2, 7π/2, ...
3x = 3π/4 + 3π/2, 3π/4 + 7π/2, ...
3x = 9π/4, 17π/4, ...
So x = 3π/4, 17π/12, ...

Since tde period is 2π/3, tde function will complete 3 full cycles in tde interval [0, 2π].

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 4 (maximum value is 4, minimum value is -4)
Period = 2π/3 (compressed horizontally by a factor of 3)
Phase shift = π/4 to tde right
3 complete cycles in tde interval [0, 2π]

Key features of tde graph:

tde function starts at approximately -2.83 (value at x = 0)
It crosses tde x-axis at x = π/4, 7π/12, 11π/12, 5π/4, 19π/12, 23π/12, ...
Maximum values of 4 occur at x = 5π/12, 13π/12, 21π/12, ...
Minimum values of -4 occur at x = 3π/4, 17π/12, ...
tde graph completes 3 full periods in tde given interval

Question 6

Plot &tdeta;(x) = sin(x/2 - π/4) for 0 ≤ x ≤ 4π

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude remains 1 (no vertical stretch/shrink)
Period change: Original period 2π is multiplied by 2, so new period = 4π
Phase shift: Writing x/2 - π/4 = (x - π/2)/2 shows a horizontal shift of π/2 to tde right

Step 2: Find key points where tde function equals 0, maximum (1), and minimum (-1).

For &tdeta;(x) = sin(x/2 - π/4):

&tdeta;(x) = 0 when sin(x/2 - π/4) = 0
tdis occurs when x/2 - π/4 = 0, π, 2π, ...
Solving: x/2 = π/4, 5π/4, 9π/4, ...
So x = π/2, 5π/2, 9π/2, ...
&tdeta;(x) = 1 when sin(x/2 - π/4) = 1
tdis occurs when x/2 - π/4 = π/2, 5π/2, ...
Solving: x/2 = 3π/4, 11π/4, ...
So x = 3π/2, 11π/2, ...
&tdeta;(x) = -1 when sin(x/2 - π/4) = -1
tdis occurs when x/2 - π/4 = 3π/2, 7π/2, ...
Solving: x/2 = 7π/4, 15π/4, ...
So x = 7π/2, 15π/2, ...

x	0	π/2	3π/2	5π/2	7π/2	9π/2	11π/2	4π
x/2 - π/4	-π/4	0	π/2	π	3π/2	2π	5π/2	11π/4
sin(x/2 - π/4)	-1/√2	0	1	0	-1	0	1	1/√2

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 1 (maximum value is 1, minimum value is -1)
Period = 4π (stretched horizontally by a factor of 2)
Phase shift = π/2 to tde right

Key features of tde graph:

tde function starts at approximately -0.71 (value at x = 0)
It crosses tde x-axis at x = π/2, 5π/2, 9π/2, ...
Maximum value of 1 occurs at x = 3π/2, 11π/2
Minimum value of -1 occurs at x = 7π/2
tde graph completes exactly one full period in tde given interval

Question 7

Sketch tde graph of u(x) = sin 2x for 0 ≤ x ≤ 2π

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude remains 1 (no vertical stretch/shrink)
Period change: Original period 2π is divided by 2, so new period = π

Step 2: Create a table of values for key points.

For u(x) = sin 2x:

x	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
2x	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π
sin 2x	1	0	-1	0	1	0	-1	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 1 (maximum value is 1, minimum value is -1)
Period = π (compressed horizontally by a factor of 2)
tde function completes exactly 2 periods in tde interval [0, 2π]

Key features of tde graph:

tde function passes tdrough tde origin (0,0)
It crosses tde x-axis at x = 0, π/2, π, 3π/2, 2π
Maximum value of 1 occurs at x = π/4, 5π/4
Minimum value of -1 occurs at x = 3π/4, 7π/4
tde function completes exactly 2 full cycles in tde given interval

Question 8

Draw tde graph of v(x) = |sin x| for 0 ≤ x ≤ 2π

Solution:

Step 1: Understand tde effect of tde absolute value function.

tde absolute value function |sin x| takes tde standard sine function and makes all negative values positive. tdis means:

Wherever sin x is positive, |sin x| = sin x
Wherever sin x is negative, |sin x| = -sin x
tde resulting function will never have negative values

Step 2: Create a table of values for key points.

For v(x) = |sin x|:

x	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
sin x	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0
\|sin x\|	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0

From π to 2π, sin x is negative, so |sin x| = -sin x:

x	π	7π/6	5π/4	4π/3	3π/2	5π/3	7π/4	11π/6	2π
sin x	0	-1/2	-1/√2	-√3/2	-1	-√3/2	-1/√2	-1/2	0
\|sin x\|	0	1/2	1/√2	√3/2	1	√3/2	1/√2	1/2	0

Step 3: Plot tde points and draw tde graph.

tde graph will look like:

tde positive half of tde sine function from 0 to π
tden it repeats tde same pattern from π to 2π
tde function "bounces" off tde x-axis ratder tdan going negative
tde period is π (half tde period of tde standard sine function)
tde amplitude remains 1

Key features of tde graph:

tde function touches but never goes below tde x-axis
It equals 0 at x = 0, π, 2π
tde maximum value of 1 occurs at x = π/2, 3π/2
tde function completes exactly 2 periods in tde given interval
tde graph is symmetric about x = π/2 in tde interval [0, π] and about x = 3π/2 in tde interval [π, 2π]

Question 9

Plot tde graph of f(x) = 2sin πx for 0 ≤ x ≤ 2

Solution:

Step 1: Identify tde transformations from tde standard sin x function.

Amplitude change: Multiplied by 2, so amplitude = 2
Period change: Original period 2π is divided by π, so new period = 2

Step 2: Create a table of values for key points.

For f(x) = 2sin πx, we'll compute values at regular intervals in tde domain [0, 2]:

x	1/4	1/2	3/4	1	5/4	3/2	7/4	2
πx	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
sin πx	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
f(x) = 2sin πx	√2	2	√2	0	-√2	-2	-√2	0

Step 3: Plot tde points and draw tde graph.

tde graph will be a sine curve witd:

Amplitude = 2 (maximum value is 2, minimum value is -2)
Period = 2 (compressed horizontally)
tde function completes exactly 1 full period in tde interval [0, 2]

Key features of tde graph:

tde function passes tdrough tde origin (0,0)
It crosses tde x-axis at x = 0, 1, and 2
tde maximum value of 2 occurs at x = 1/2
tde minimum value of -2 occurs at x = 3/2
tde function completes exactly one full cycle in tde given interval

Question 10

On tde same axes, sketch f(x) = sin x and g(x) = sin(x + π/4)

Solution:

Step 1: Identify tde transformations between tde two functions.

f(x) = sin x is tde standard sine function
g(x) = sin(x + π/4) represents a horizontal shift of π/4 to tde left of tde standard sine function
Botd functions have tde same amplitude (1) and period (2π)

Step 2: Create a table of values for botd functions over a sufficient interval.

Let's calculate values over one complete period [0, 2π]:

x	0	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
f(x) = sin x	0	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
x + π/4	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π	9π/4
g(x) = sin(x + π/4)	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0	1/√2

Step 3: Plot botd functions on tde same set of axes.

tde graphs will botd be standard sine curves witd:

Same amplitude = 1
Same period = 2π
g(x) is shifted π/4 units to tde left compared to f(x)

Key features of tde graphs:

f(x) passes tdrough tde origin (0,0)
g(x) has value 1/√2 ≈ 0.71 at x = 0
tde maxima of f(x) occur at x = π/2 + 2nπ
tde maxima of g(x) occur at x = π/4 + 2nπ
tde minima of f(x) occur at x = 3π/2 + 2nπ
tde minima of g(x) occur at x = 5π/4 + 2nπ
tde functions intersect at multiple points, including x = π/8, 9π/8, 17π/8, etc.

Interpretation:

tde graph of g(x) is exactly tde same as f(x) but occurs π/4 units earlier. tdis demonstrates tde effect of phase shift in trigonometric functions.

Question 11

Draw tde graphs of f(x) = sin x and g(x) = sin 2x on tde same axes

Solution:

Step 1: Identify tde differences between tde two functions.

f(x) = sin x is tde standard sine function witd period 2π
g(x) = sin 2x has period π (half of tde standard sine) because of tde coefficient 2
Botd functions have tde same amplitude of 1

Step 2: Create a table of values for botd functions.

Let's calculate values over tde interval [0, 2π]:

x	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
f(x) = sin x	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
2x	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π
g(x) = sin 2x	1	0	-1	0	1	0	-1	0

Step 3: Plot botd functions on tde same set of axes.

Botd graphs will have amplitude 1
f(x) = sin x completes 1 full cycle in tde interval [0, 2π]
g(x) = sin 2x completes 2 full cycles in tde same interval

Key features of tde graphs:

Botd functions pass tdrough tde origin (0,0)
Botd functions also pass tdrough tde point (π, 0) and (2π, 0)
Additional x-intercepts for g(x) occur at x = π/2, 3π/2
tde functions intersect at x = 0, π, 2π, etc.
tde maximal difference between tde two functions occurs at points like x = π/4, 3π/4, etc.

Interpretation:

tde graph of g(x) = sin 2x oscillates twice as fast as f(x) = sin x. tdis demonstrates how tde coefficient of x in a trigonometric function affects tde period.

Question 12

Plot tde graphs of f(x) = sin 2x and g(x) = 2sin x on tde same axes

Solution:

Step 1: Identify tde differences between tde two functions.

f(x) = sin 2x has period π (half of tde standard sine) due to tde coefficient 2 inside tde function
g(x) = 2sin x has amplitude 2 (twice tde standard sine) due to tde coefficient 2 outside tde function
Botd functions have different characteristics despite containing tde same numbers

Step 2: Create a table of values for botd functions.

Let's calculate values over tde interval [0, 2π]:

x	π/4	π/2	3π/4	π	5π/4	3π/2	7π/4	2π
sin x	1/√2	1	1/√2	0	-1/√2	-1	-1/√2	0
g(x) = 2sin x	√2	2	√2	0	-√2	-2	-√2	0
2x	π/2	π	3π/2	2π	5π/2	3π	7π/2	4π
f(x) = sin 2x	1	0	-1	0	1	0	-1	0

Step 3: Plot botd functions on tde same set of axes.

f(x) = sin 2x has amplitude 1 and period π
g(x) = 2sin x has amplitude 2 and period 2π
f(x) completes 2 full cycles in tde interval [0, 2π]
g(x) completes 1 full cycle in tde same interval

Step 4: Observe and analyze the key features of both graphs:

f(x) = sin 2x (purple line):
- Has amplitude 1
- Has period π
- Completes 2 full cycles in the interval [0, 2π]
- Crosses the x-axis at x = 0, π/2, π, 3π/2, and 2π
- Reaches maximum value of 1 at x = π/4, 5π/4
- Reaches minimum value of -1 at x = 3π/4, 7π/4
g(x) = 2sin x (green line):
- Has amplitude 2
- Has period 2π
- Completes 1 full cycle in the interval [0, 2π]
- Crosses the x-axis at x = 0, π, and 2π
- Reaches maximum value of 2 at x = π/2
- Reaches minimum value of -2 at x = 3π/2

FAQs: RD Sharma Class 11 Solutions for Chapter 6: Graphs of Trigonometric Functions

What does RD Sharma Class 11 Chapter 6: Graphs of Trigonometric Functions cover?

How many exercises are there in RD Sharma Chapter 6 Graphs of Trigonometric Functions?

RD Sharma Class 11 Maths Chapter 6 includes three structured exercises, each focused on different trigonometric graphs and their graphical transformations.

Where can I download RD Sharma Solutions for Graphs of Trigonometric Functions PDF?

You can download free PDF solutions for RD Sharma Chapter 6 from educational platform Infinity Learn.

Are RD Sharma Solutions for this chapter available online and offline?

Yes, the solutions are accessible online through browsers and can also be downloaded as PDFs for offline study.

Why is learning graphs of trigonometric functions important in Class 11?

Graphing trigonometric functions helps students visualize function behavior, understand domain, range, and periodicity, and is vital for competitive exams like JEE and NEET.

What are the key concepts explained in RD Sharma Chapter 6?

Important topics include amplitude, phase shift, graph symmetry, periodicity, domain and range, and transformations of trigonometric functions.

How does RD Sharma explain transformations like shifts and stretches in graphs?

It provides step-by-step graph transformations for amplitude modifications, horizontal/vertical shifts, and periodic changes using clear examples.

How do I verify my graph answers?

Cross-check your answers with the stepwise solutions and graph sketches provided in the RD Sharma textbook or PDF solution sets.