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NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.2

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Get NCERT Solutions for Maths Class 11 Chapter 3 Exercise 3.2 - FREE PDF Download

Trigonometric functions in NCERT Class 11 Maths Chapter 3 Exercise 3.2 introduce students to the fundamental relationships between angles and sides in triangles. These functions, including sine, cosine, and tangent, are essential for understanding geometric properties and solving practical problems in physics, engineering, and architecture. 

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Table of Content
1. Get NCERT Solutions for Maths Class 11 Chapter 3 Exercise 3.2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 3 Exercise 3.2 Class 11 | Vedantu
3. Access NCERT Solutions for Maths Class 11 Chapter 3 - Trigonometric Functions
    3.1Exercise 3.2
    3.2Points to Remember Before Solving Exercise 3.2
4. Class 11 Maths Chapter 3: Exercises Breakdown
5. CBSE Class 11 Maths Chapter 3 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs


Students will focus on learning how to apply trigonometric ratios to calculate angles and side lengths in right-angled triangles. Mastering these functions lays the groundwork for advanced topics such as trigonometric identities and equations, which are crucial for higher studies in mathematics and related disciplines. The NCERT Solutions for Class 11 Maths provided by Vedantu offer comprehensive explanations and exercises to ensure a thorough understanding of these concepts per the updated CBSE Syllabus for Class 11 Mathematics.


Glance on NCERT Solutions Maths Chapter 3 Exercise 3.2 Class 11 | Vedantu

  • Exercise 3.2 Class 11 Maths explains the sign convention of trigonometric functions, which determines whether the trigonometric ratios are positive or negative based on the quadrant of the angle in the standard position.

  • It covers the domain of trigonometric functions, which consists of all possible input values (angles) for which the functions are defined. The range of trigonometric functions represents all possible output values (ratios) that the functions can produce.

  • It discusses the graphical representation of trigonometric functions sine (sin), cosine (cos), tangent (tan), cosecant (CSC), secant (sec), and cotangent (cot) on a coordinate plane. This illustrates the periodic nature of sine and cosine functions and their amplitude and period.

  • It explains how the sign convention affects the values of trigonometric functions in different quadrants.

  • Finally, it includes examples and exercises to reinforce understanding of the sign convention, domain, range, and graphical representation of trigonometric functions.

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Access NCERT Solutions for Maths Class 11 Chapter 3 - Trigonometric Functions

Exercise 3.2

1. Find the values of the other five trigonometric functions if  $\text{cos x=-}\frac{\text{1}}{\text{2}}$ , $x$ lies in the third quadrant.

Ans:

Here given that,  $\text{cos x=-}\frac{\text{1}}{\text{2}}$

Therefore we have,

$\text{sec x=}\frac{\text{1}}{\text{cos x}}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}$

$\text{=-2}$

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{si}{{\text{n}}^{\text{2}}}\text{x=1-co}{{\text{s}}^{\text{2}}}\text{x}$

Substituting  $\text{cos x=-}\frac{\text{1}}{\text{2}}$ in the formula, we obtain,

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}{{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}\frac{\text{1}}{\text{4}}$

$\text{=}\frac{\text{3}}{\text{4}}$

$\text{sin x= }\!\!\pm\!\!\text{ }\frac{\sqrt{\text{3}}}{\text{2}}$

Since $\text{x}$ lies in the ${{\text{3}}^{\text{rd}}}$quadrant, the value of $\sin x$ will be negative.

$\text{sin x=-}\frac{\sqrt{\text{3}}}{\text{2}}$

Therefore, $\text{cosec x=}\frac{\text{1}}{\text{sin x}}$ 

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\sqrt{\text{3}}}{\text{2}} \right)}$ 

$\text{=-}\frac{\text{2}}{\sqrt{\text{3}}}$ 

Hence ,

$\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

$\text{=}\frac{\left( \text{-}\frac{\sqrt{\text{3}}}{\text{2}} \right)}{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}$

$\text{=}\sqrt{\text{3}}$

And 

$\text{cot x=}\frac{\text{1}}{\text{tan x}}$

$\text{=}\frac{\text{1}}{\sqrt{\text{3}}}$


2. Find the values of other five trigonometric functions if $\text{sin  x=}\frac{\text{3}}{\text{5}}$  , $\text{x}$ lies in second quadrant.

Ans:

Here given that,  $\text{sin x=}\frac{\text{3}}{\text{5}}$

Therefore we have,

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$=\frac{1}{\left( \frac{3}{5} \right)}$

$=\frac{5}{3}$

Now we know that , ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Therefore we have, $\text{co}{{\text{s}}^{\text{2}}}\text{x=1-si}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\sin x=\frac{3}{5}$  in the formula, we obtain,

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}$

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}\frac{\text{9}}{\text{25}}$

$\text{=}\frac{\text{16}}{\text{25}}$ 

$\text{cos x= }\!\!\pm\!\!\text{ }\frac{\text{4}}{\text{5}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\cos x$ will be negative.

$\text{cos x=-}\frac{\text{4}}{\text{5}}$

Therefore, $sec x=\frac{1}{\cos x}$ 

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{4}}{\text{5}} \right)}$ 

$\text{=-}\frac{\text{5}}{\text{4}}$ 

Hence ,

$\tan x=\frac{\sin  x}{\cos x}$ 

$\text{=}\frac{\left( \frac{\text{3}}{\text{5}} \right)}{\left( \text{-}\frac{\text{4}}{\text{5}} \right)}$

$\text{=-}\frac{\text{3}}{\text{4}}$

And 

$\cot x=\frac{1}{\tan x}$

$\text{=-}\frac{\text{4}}{\text{3}}$


3. Find the values of other five trigonometric functions if $\text{cot x=}\frac{\text{3}}{\text{4}}$ , $\text{x}$ lies in third quadrant.

Ans:

Here given that,  $\cot x=\frac{3}{4}$

Therefore we have,

$\tan x=\frac{1}{\cot x}$

$=\frac{1}{\left( \frac{3}{4} \right)}$

$=\frac{4}{3}$

Now we know that , \[\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}\]

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=}\frac{\text{4}}{\text{3}}$  in the formula, we obtain,

${{\sec }^{2}}x=1+{{\left( \frac{4}{3} \right)}^{2}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\frac{\text{16}}{\text{9}}$

$=\frac{25}{9}$ 

$\text{sec x= }\!\!\pm\!\!\text{ }\frac{\text{5}}{\text{3}}$

Since $x$ lies in the ${{3}^{rd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\frac{\text{5}}{\text{3}}$

Therefore, $\cos  x=\frac{1}{\sec x}$ 

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{5}}{\text{3}} \right)}$ 

$\text{=-}\frac{\text{3}}{\text{5}}$ 

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

Hence we have, $\text{sin x=}\frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{3}}{\text{5}} \right)$   

\[\text{=}\left( \text{-}\frac{\text{4}}{\text{5}} \right)\] 

And 

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=-}\frac{\text{5}}{\text{4}}$


4. Find the values of other five trigonometric functions if $\text{sec  x=}\frac{\text{13}}{\text{5}}$  , $\text{x}$ lies in fourth quadrant.

Ans:

Here given that,  $\sec x=\frac{13}{5}$

Therefore we have,

$\cos x=\frac{1}{\sec x}$

$=\frac{1}{\left( \frac{13}{5} \right)}$

$=\frac{5}{13}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{ta}{{\text{n}}^{\text{2}}}\text{x=se}{{\text{c}}^{\text{2}}}\text{x-1}$

Substituting  \[\text{sec x=}\frac{\text{13}}{\text{5}}\]  in the formula, we obtain,

\[\text{ta}{{\text{n}}^{\text{2}}}\text{x=}{{\left( \frac{\text{13}}{\text{5}} \right)}^{\text{2}}}\text{-1}\]

$\text{ta}{{\text{n}}^{\text{2}}}\text{x=}\frac{\text{169}}{\text{25}}\text{-1}$

$\text{=}\frac{\text{144}}{\text{25}}$ 

\[\text{tanx= }\!\!\pm\!\!\text{ }\frac{\text{12}}{\text{5}}\]

Since $x$ lies in the ${{4}^{th}}$ quadrant, the value of $\tan x$ will be negative.

$\text{tan x=-}\frac{\text{12}}{\text{5}}$

Therefore, \[\text{cot x=}\frac{\text{1}}{\text{tan x}}\] 

$\text{=-}\frac{\text{5}}{\text{12}}$ 

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

Hence we have, $\text{sin x=}\frac{\text{5}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{12}}{\text{5}} \right)$   

$\text{=}\left( \text{-}\frac{\text{12}}{\text{13}} \right)$ 

And 

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=-}\frac{\text{13}}{\text{12}}$


5. Find the values of other five trigonometric functions if  $\text{tan x=-}\frac{\text{5}}{\text{12}}$ , $\text{x}$ lies in second quadrant.

Ans:

Here given that,  $\text{tan x=-}\frac{\text{5}}{\text{12}}$

Therefore we have,

$\text{cot x=}\frac{\text{1}}{\text{tan x}}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{5}}{\text{12}} \right)}$

$\text{=-}\frac{\text{12}}{\text{5}}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=-}\frac{\text{5}}{\text{12}}$  in the formula, we obtain,

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}{{\left( \text{-}\frac{\text{5}}{\text{12}} \right)}^{\text{2}}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\frac{\text{25}}{\text{144}}$

$=\frac{169}{144}$ 

$\text{sec x= }\!\!\pm\!\!\text{ }\frac{\text{13}}{\text{12}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\frac{\text{13}}{\text{12}}$

Therefore, $\text{cos x=}\frac{\text{1}}{\text{sec x}}$ 

$\text{=-}\frac{\text{12}}{\text{13}}$ 

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

Hence we have, $\text{sin x=}\left( \text{-}\frac{\text{5}}{\text{12}} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{12}}{\text{13}} \right)$   

$=\left( \frac{5}{13} \right)$ 

And 

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=}\frac{\text{13}}{\text{5}}$


6. Find the value of the trigonometric function $\text{sin76}{{\text{5}}^{\text{o}}}$ .

Ans: We know that the values of $\sin x$ repeat after an interval of $2\pi $ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin76}{{\text{5}}^{\text{o}}}\text{=sin}\left( \text{2 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}}\text{+4}{{\text{5}}^{\text{o}}} \right)$

$\text{=sin4}{{\text{5}}^{\text{o}}}$

$\text{=}\frac{\text{1}}{\sqrt{\text{2}}}\text{.}$


7. Find the value of the trigonometric function $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)$  

Ans: We  know that the values of $\text{cosec x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+4 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}} \right)$

$\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+144}{{\text{0}}^{\text{o}}} \right)$

$\text{=cosec3}{{\text{0}}^{\text{o}}}$

$=2$ 


8. Find the value of the trigonometric function  $\text{tan}\frac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}$ .

Ans: We know that the values of $\text{tan x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

Therefore we can write,

$\text{tan}\frac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=tan6}\frac{\text{1}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$

$\text{=tan}\left( \text{6 }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

\[\text{=tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

$\text{=}\sqrt{\text{3}}$      


9. Find the value of the trigonometric function $\text{sin}\left( \text{-}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$ 

Ans: We know that the values of $\text{sin x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin}\left( \text{-}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=sin}\left( \text{-}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}}\text{+2 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ } \right)$ 

$\text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

$=\frac{\sqrt{3}}{2}$


10. Find the value of the trigonometric function $\text{cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

Ans: We know that the values of $\text{cot x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

Therefore we can write,

$\text{cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+4 }\!\!\pi\!\!\text{ } \right)$ 

$\text{=cot}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

$=1$


Points to Remember Before Solving Exercise 3.2

1. Trigonometric Identities 

  • sin2 A + cos2 A = 1

  • 1 + tan2 A = sec2 A

  • 1 + cot2 A = cosec2 A


2. Domain and Range of Trigonometric Functions

The input and output values of trigonometric functions, respectively, are the domain and range of trigonometric functions. The range of trigonometric functions denotes the resultant value of the trigonometric function corresponding to a given angle in the domain, whereas the domain of trigonometric functions denotes the values of angles where the trigonometric functions are given.


3. Sign of Trigonometric Functions in Different Quadrants

All functions are positive in the first quadrant, only sin and cosec are positive in the second quadrant, and only tan and cot are positive in the third quadrant, and only cos and sec are positive in the fourth quadrant. 


4. Domain and Range Table for Trigonometric Functions

Trigonometric Functions

Domain

Range

sin θ

(-∞, + ∞)

[-1, +1]

cos θ

(-∞, +∞)

[-1, +1]

tan θ

R - (2n + 1)π/2

(-∞, +∞)

cot θ

R - nπ

(-∞, +∞)

sec θ

R - (2n + 1)π/2

(-∞, -1] U [+1, +∞)

cosec θ

R - nπ

(-∞, -1] U [+1, +∞)


Conclusion

Understanding trigonometric functions is crucial for solving problems involving angles and triangles. Focus on mastering the definitions, sign conventions, domain, range, and graphical representations of sine, cosine, and tangent functions. These concepts are foundational in fields like engineering, physics, and architecture. Previous year question papers typically include 2-3 questions on trigonometric functions, highlighting their importance in exams. Practice diverse problems and use NCERT Solutions to strengthen your understanding and preparation. Clear concepts in trigonometry will not only aid in academic success but also in practical applications in various disciplines.


Class 11 Maths Chapter 3: Exercises Breakdown

Exercise

Number of Questions

Exercise 3.1

7 Questions and Solutions

Exercise 3.3

25 Questions and Solutions

Miscellaneous Exercise

10 Questions and Solutions


CBSE Class 11 Maths Chapter 3 Other Study Materials


Chapter-Specific NCERT Solutions for Class 11 Maths

The chapter-wise NCERT Solutions for Class 11 Maths are given below. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Related Links for CBSE Class 11 Maths

FAQs on NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.2

1. What are the trigonometric functions taught in Class 11 Maths Ex 3.2?

Trigonometric functions relate the angles of a triangle to ratios of its sides, like sine, cosine, and tangent. Trigonometric functions are fundamental in mathematics and have wide-ranging applications. Understanding their concepts and applications will help in both academic success and practical problem-solving.

2. Why are sign conventions important for ex 3.2 class 11 in trigonometry?

Sign conventions determine whether trigonometric ratios are positive or negative based on the quadrant of the angle.

3. How do you find the domain of trigonometric functions for class 11 ex 3.2?

The domain of trigonometric functions includes all angles where the functions are defined.

4. What is the range of trigonometric functions in exercise 3.2 class 11 maths solutions?

The range of trigonometric functions includes all possible values (ratios) that the functions can produce.

5. How do exercise 3.2 class 11 maths solutions graphically represent trigonometric functions?

Trigonometric functions are graphed on a coordinate plane, showing their periodic nature, amplitude, and period.

6. What real-world applications of trigonometric functions exercise 3.2 class 11?

Trigonometric functions are used in navigation, physics, engineering, and architecture to calculate angles and distances.

7. What are reciprocal trigonometric functions taught in exercise 3.2, class 11 maths solutions?

Reciprocal trigonometric functions include cosecant, secant, and cotangent, which are inverses of sine, cosine, and tangent functions.

8. How many questions on trigonometric functions in class 11 maths chapter 3 exercise 3.2 were asked in previous year exams?

Typically, previous year question papers include 2-3 questions on trigonometric functions.

9. What should students focus on while going through trigonometric functions class 11 maths exercise 3.2 solutions?

Focus on understanding definitions, sign conventions, domain, range, graphical representation, and applications of trigonometric functions.

10. How do trigonometric functions help solve triangles in class 11 maths 3.2 exercise?

Trigonometric functions help calculate the angles and sides of triangles, which is useful in solving geometric problems.