RS Aggarwal Solutions for Class 11 Chapter 19 Graphs Of Trigonometric

Access free RS Aggarwal Solutions for Class 11 Chapter 19 Graphs Of Trigonometric 2026 below. Students can now access free RS Aggarwal Solutions Solutions for Class 11 Mathematics. These chapter-wise exercises are designed by expert math teachers to help you understand complex formulas and score higher marks in your class tests.

Class 11 Math Chapter 19 Graphs Of Trigonometric RS Aggarwal Solutions Solutions

Get step-by-step RS Aggarwal Solutions Solutions for Chapter 19 Graphs Of Trigonometric Class 11 Math below. All answers are updated for the 2026 school curriculum, offering step by step methods to help you solve textbook problems easily.

Chapter 19 Graphs Of Trigonometric RS Aggarwal Solutions Class 11 Solved Exercises

 

Question 1. Draw the graph of each of the following functions: Sin 3x
Answer: To sketch the curve sin(3x), select certain standard angle values that help you pinpoint the key points and sketch the curve.

The value table is:

x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
sin(3x)10-1010

The resulting graph shows the curve sin(3x). In this case, the oscillation rate of the function sin(x) is boosted 3 times over.
In simple words: When you multiply the input by 3, the wave squeezes horizontally and completes more cycles in the same space. The height stays the same - only the width shrinks.

Exam Tip: Always identify amplitude (height) and period (width) separately - amplitude stays 1, but period changes from 2π to 2π/3, making the function repeat 3 times faster.

 

Question 2. Draw the graph of each of the following functions: 3sin x
Answer: To sketch the curve 3sin(x), select certain standard angle values that help you locate the key points and sketch the curve.

The value table is:

x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
3sin(x)\( \frac{3}{2} \)\( \frac{3\sqrt{3}}{2} \)30-30

The resulting graph shows the curve 3sin(x). In this case, the peak height of the function sin(x) is stretched vertically 3 times.
In simple words: When you multiply the whole function by 3, the wave gets taller - it now swings from -3 to 3 instead of -1 to 1. The wave still completes one cycle in the same 2π distance, but it stretches upward and downward.

Exam Tip: Amplitude (vertical stretch) changes but period remains 2π - identify both separately to avoid confusion between scaling the input versus scaling the output.

 

Question 3. Draw the graph of each of the following functions: 2sin 3x
Answer: To sketch the curve 2sin(3x), select certain standard angle values that help you locate the key points and sketch the curve.

The value table is:

x\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
2sin(3x)2020

The resulting curve combines both effects: the wave cycles 3 times faster and stretches twice as tall as the base sine function.
In simple words: This function has two changes applied at once - it bounces up to 2 and down to -2 (taller), and completes 3 full waves in the space where sin(x) does just one. Both transformations happen together.

Exam Tip: Always separate amplitude changes from frequency changes - amplitude is the multiplier in front (here 2), and frequency depends on the coefficient of x (here 3).

 

Question 4. Draw the graph of each of the following functions: 2cos 3x
Answer: To sketch the curve 2cos(3x), select certain standard angle values that help you locate the key points and sketch the curve.

The value table is:

x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
2cos(3x)0-20-202

The resulting curve shows that both the wave's vertical height and its oscillation rate have increased - the amplitude of the cosine function grows 2 times and the oscillation speed grows 3 times.
In simple words: This combines a taller wave (amplitude 2 instead of 1) with a faster cycle (3 times faster). The cosine starts at its peak of 2 and moves through its pattern more quickly than the standard cos(x).

Exam Tip: The cosine function starts at a maximum, unlike sine which starts at zero - use this feature to sketch the starting point correctly on your graph.

 

Question 5. Draw the graph of each of the following functions: \( \sin \frac{x}{2} \)
Answer: To sketch the curve sin(x/2), select certain standard angle values that help you locate the key points and sketch the curve.

The value table is:

x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
sin(x/2)\( \frac{\sqrt{3} - 1}{2\sqrt{2}} \)\( \frac{1}{2} \)\( \frac{1}{\sqrt{2}} \)1\( \frac{1}{\sqrt{2}} \)0

The resulting curve shows that the oscillation rate of the sine function has been cut in half. In other words, the wave now moves twice as slowly as the base sin(x) function.
In simple words: Dividing x by 2 makes the wave stretch out horizontally. Instead of completing one cycle in 2π, the function now needs 4π to finish one complete wave. Everything happens at half the normal speed.

Exam Tip: When the coefficient of x is a fraction (less than 1), the period grows - always compute the new period as 2π divided by the coefficient to find how stretched the wave becomes.

 

Question 6. Draw the graphs of y = sin x and \( y = \cos x \) in [0, 2\pi] on the same axes.
Answer: For sin(x), the value table is:

x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
sin(x)\( \frac{1}{2} \)\( \frac{\sqrt{3}}{2} \)10-10

For cos(x), the value table is:
x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
cos(x)\( \frac{\sqrt{3}}{2} \)\( \frac{1}{2} \)0-101

When both curves are plotted together on the same coordinate system, the green line shows sin(x) and the blue line shows cos(x) over the interval [0, 2π]. The cosine curve appears shifted to the left of the sine curve by π/2 radians.
In simple words: Both waves have the same height (amplitude 1) and take the same time to repeat (period 2π), but they start at different points. Cosine starts at its peak of 1, while sine begins at 0. The cosine is ahead by one-quarter of a full cycle.

Exam Tip: Remember the key property: cos(x) = sin(x + π/2) - the cosine graph is the sine graph shifted π/2 units to the left, which helps you sketch both quickly.

 

Question 7. Draw the graphs of y = cos x and \( y = \cos 2x \) in [0, 2\pi] on the same axes.
Answer: For cos(x), the value table is:

x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
cos(x)\( \frac{\sqrt{3}}{2} \)\( \frac{1}{2} \)0-101

For cos(2x), the value table is:
x\( \frac{\pi}{6} \)\( \frac{\pi}{3} \)\( \frac{\pi}{2} \)\( \pi \)\( \frac{3\pi}{2} \)\( 2\pi \)
cos(2x)\( \frac{1}{2} \)\( -\frac{1}{2} \)-11-11

When both curves are plotted together, the blue line represents cos(2x) and the purple line represents cos(x) over [0, 2π]. The cos(2x) curve oscillates twice as quickly - it completes 2 full cycles while cos(x) completes just 1 cycle in the same interval. Both begin at their maximum value of 1, but cos(2x) reaches its extrema much faster.
In simple words: Both start at 1, but cos(2x) oscillates twice as fast as cos(x). In the time cos(x) makes one complete wave, cos(2x) finishes two waves. The period of cos(2x) is π instead of 2π.

Exam Tip: The period formula is 2π/|coefficient of x| - use this to predict how many cycles appear in a given interval before you plot any points.

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Yes, practicing these exercises thoroughly will significantly improve your foundational concepts. The step-by-step layout helps you understand how formulas are applied, ensuring you score top marks in your Class 11 tests and school examinations.

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