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| Boilerpipe Text | NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
are available at BYJU’S, which are prepared by our expert teachers. All these solutions are written as per the latest update on the CBSE Syllabus 2023-24 and its guidelines. BYJU’S provides step-by-step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3, Trigonometric Functions, comes under NCERT Class 11 Maths and is an important chapter for students. Though the chapter has numerous mathematical terms and formulae, BYJU’S has made
NCERT Solutions for Class 11 Maths
easy for the students to understand and remember with the usage of tricks.
Trigonometry has been developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is used in many areas, such as finding the heights of tides in the ocean, designing electronic circuits, etc. In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In
Class 11
, trigonometric ratios are generalised to trigonometric function and their properties. However, the
NCERT Solutions
of BYJU’S help the students to attain more knowledge and score full marks in this chapter of the examination.
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
Download PDF
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In this section, a few important terms are defined, such as principal solution and general solution of trigonometric functions, which are explained using examples too.
Exercise 3.1 Solutions
7 Questions
Exercise 3.2 Solutions
10 Questions
Exercise 3.3 Solutions
25 Questions
Exercise 3.4 Solutions
9 Questions
Miscellaneous Exercise On Chapter 3 Solutions
10 Questions
Access NCERT Solutions for Class 11 Maths Chapter 3
Exercise 3.1 page: 54
1. Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°
Solution:
(iv) 520°
2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)
(i) 11/16
(ii) -4
(iii) 5π/3
(iv) 7π/6
Solution:
(i) 11/16
Here π radian = 180°
(ii) -4
Here π radian = 180°
(iii) 5π/3
Here π radian = 180°
We get
= 300
o
(iv) 7π/6
Here π radian = 180°
We get
= 210
o
3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Solution:
It is given that
No. of revolutions made by the wheel in
1 minute = 360
1 second = 360/60 = 6
We know that
The wheel turns an angle of 2π radian in one complete revolution.
In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian
Therefore, in one second, the wheel turns an angle of 12π radian.
4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).
Solution:
5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Solution:
The dimensions of the circle are
Diameter = 40 cm
Radius = 40/2 = 20 cm
Consider AB be as the chord of the circle i.e. length = 20 cm
In ΔOAB,
Radius of circle = OA = OB = 20 cm
Similarly AB = 20 cm
Hence, ΔOAB is an equilateral triangle.
θ = 60° = π/3 radian
In a circle of radius
r
unit, if an arc of length
l
unit subtends an angle
θ
radian at the centre
We get θ = 1/r
Therefore, the length of the minor arc of the chord is 20π/3 cm.
6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Solution:
7. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm (ii) 15 cm (iii) 21 cm
Solution:
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r
We know that r = 75 cm
(i) l = 10 cm
So we get
θ = 10/75 radian
By further simplification
θ = 2/15 radian
(ii) l = 15 cm
So we get
θ = 15/75 radian
By further simplification
θ = 1/5 radian
(iii) l = 21 cm
So we get
θ = 21/75 radian
By further simplification
θ = 7/25 radian
Exercise 3.2 page: 63
Find the values of other five trigonometric functions in Exercises 1 to 5.
1. cos x = -1/2, x lies in third quadrant.
Solution:
2. sin x = 3/5, x lies in second quadrant.
Solution:
It is given that
sin x = 3/5
We can write it as
We know that
sin
2
x + cos
2
x = 1
We can write it as
cos
2
x = 1 – sin
2
x
3. cot x = 3/4, x lies in third quadrant.
Solution:
It is given that
cot x = 3/4
We can write it as
We know that
1 + tan
2
x = sec
2
x
We can write it as
1 + (4/3)
2
= sec
2
x
Substituting the values
1 + 16/9 = sec
2
x
cos
2
x = 25/9
sec x = ± 5/3
Here x lies in the third quadrant so the value of sec x will be negative
sec x = – 5/3
We can write it as
4. sec x = 13/5, x lies in fourth quadrant.
Solution:
It is given that
sec x = 13/5
We can write it as
We know that
sin
2
x + cos
2
x = 1
We can write it as
sin
2
x = 1 – cos
2
x
Substituting the values
sin
2
x = 1 – (5/13)
2
sin
2
x = 1 – 25/169 = 144/169
sin
2
x = ± 12/13
Here x lies in the fourth quadrant so the value of sin x will be negative
sin x = – 12/13
We can write it as
5. tan x = -5/12, x lies in second quadrant.
Solution:
It is given that
tan x = – 5/12
We can write it as
We know that
1 + tan
2
x = sec
2
x
We can write it as
1 + (-5/12)
2
= sec
2
x
Substituting the values
1 + 25/144 = sec
2
x
sec
2
x = 169/144
sec x = ± 13/12
Here x lies in the second quadrant so the value of sec x will be negative
sec x = – 13/12
We can write it as
Find the values of the trigonometric functions in Exercises 6 to 10.
6. sin 765°
Solution:
We know that values of sin x repeat after an interval of 2π or 360°
So we get
By further calculation
= sin 45
o
= 1/
√
2
7. cosec (–1410°)
Solution:
We know that values of cosec x repeat after an interval of 2π or 360°
So we get
By further calculation
= cosec 30
o
= 2
8.
Solution:
We know that values of tan x repeat after an interval of π or 180°
So we get
By further calculation
We get
= tan 60
o
=
√
3
9.
Solution:
We know that values of sin x repeat after an interval of 2π or 360°
So we get
By further calculation
10.
Solution:
We know that values of tan x repeat after an interval of π or 180°
So we get
By further calculation
Exercise 3.3 page: 73
Prove that:
1.
Solution:
2.
Solution:
Here
= 1/2 + 4/4
= 1/2 + 1
= 3/2
= RHS
3.
Solution:
4.
Solution:
5. Find the value of:
(i) sin 75
o
(ii) tan 15
o
Solution:
(ii) tan 15°
It can be written as
= tan (45° – 30°)
Using formula
Prove the following:
6.
Solution:
7.
Solution:
8.
Solution:
9.
Solution:
Consider
It can be written as
= sin x cos x (tan x + cot x)
So we get
10. sin (
n
+ 1)
x
sin (
n
+ 2)
x
+ cos (
n
+ 1)
x
cos (
n
+ 2)
x
= cos
x
Solution:
LHS = sin (
n
+ 1)
x
sin (
n
+ 2)
x
+ cos (
n
+ 1)
x
cos (
n
+ 2)
x
11.
Solution:
Consider
Using the formula
12. sin
2
6
x
– sin
2
4
x
= sin 2
x
sin 10
x
Solution:
13. cos
2
2
x
– cos
2
6
x
= sin 4
x
sin 8
x
Solution:
We get
= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]
It can be written as
= [2 cos 4
x
cos 2
x
] [–2 sin 4
x
(–sin 2
x
)]
So we get
= (2 sin 4
x
cos 4
x
) (2 sin 2
x
cos 2
x
)
= sin 8x sin 4x
= RHS
14. sin 2x + 2sin 4x + sin 6x = 4cos
2
x sin 4x
Solution:
By further simplification
= 2 sin 4x cos (– 2x) + 2 sin 4x
It can be written as
= 2 sin 4x cos 2x + 2 sin 4x
Taking common terms
= 2 sin 4x (cos 2x + 1)
Using the formula
= 2 sin 4x (2 cos
2
x – 1 + 1)
We get
= 2 sin 4x (2 cos
2
x)
= 4cos
2
x sin 4x
= R.H.S.
15. cot 4
x
(sin 5
x
+ sin 3
x
) = cot
x
(sin 5
x
– sin 3
x
)
Solution:
Consider
LHS = cot 4x (sin 5x + sin 3x)
It can be written as
Using the formula
= 2 cos 4x cos x
Hence, LHS = RHS.
16.
Solution:
Consider
Using the formula
17.
Solution:
18.
Solution:
19.
Solution:
20.
Solution:
21.
Solution:
22. cot
x
cot 2
x
– cot 2
x
cot 3
x
– cot 3
x
cot
x
= 1
Solution:
23.
Solution:
Consider
LHS = tan 4x = tan 2(2x)
By using the formula
24. cos 4
x
= 1 – 8sin
2
x
cos
2
x
Solution:
Consider
LHS = cos 4x
We can write it as
= cos 2(2
x
)
Using the formula cos 2
A
= 1 – 2 sin
2
A
= 1 – 2 sin
2
2
x
Again by using the formula sin2
A
= 2sin
A
cos
A
= 1 – 2(2 sin
x
cos
x
)
2
So we get
= 1 – 8 sin
2
x
cos
2
x
= R.H.S.
25. cos 6
x
= 32 cos
6
x
– 48 cos
4
x
+ 18 cos
2
x
– 1
Solution:
Consider
L.H.S. = cos 6
x
It can be written as
= cos 3(2
x
)
Using the formula cos 3
A
= 4 cos
3
A
– 3 cos
A
= 4 cos
3
2
x
– 3 cos
2
x
Again by using formula cos 2
x
= 2 cos
2
x
– 1
= 4 [(2 cos
2
x
– 1)
3
– 3 (2 cos
2
x
– 1)
By further simplification
= 4 [(2 cos
2
x
)
3
– (1)
3
– 3 (2 cos
2
x
)
2
+ 3 (2 cos
2
x
)] – 6cos
2
x
+ 3
We get
= 4 [8cos
6
x
– 1 – 12 cos
4
x
+ 6 cos
2
x
] – 6 cos
2
x
+ 3
By multiplication
= 32 cos
6
x
– 4 – 48 cos
4
x
+ 24 cos
2
x
– 6 cos
2
x
+ 3
On further calculation
= 32 cos
6
x
– 48 cos
4
x
+ 18 cos
2
x
– 1
= R.H.S.
Exercise 3.4 PAGE: 78
Find the principal and general solutions of the following equations:
1. tan x = √3
Solution:
2. sec x = 2
Solution:
3. cot x = – √3
Solution:
4. cosec x = – 2
Solution:
Find the general solution for each of the following equations:
5. cos 4x = cos 2x
Solution:
6. cos 3x + cos x – cos 2x = 0
Solution:
7. sin 2x + cos x = 0
Solution:
It is given that
sin 2x + cos x = 0
We can write it as
2 sin x cos x + cos x = 0
cos x (2 sin x + 1) = 0
cos x = 0 or 2 sin x + 1 = 0
Let cos x = 0
8. sec
2
2x = 1 – tan 2x
Solution:
It is given that
sec
2
2x = 1 – tan 2x
We can write it as
1 + tan
2
2x = 1 – tan 2x
tan
2
2x + tan 2x = 0
Taking common terms
tan 2x (tan 2x + 1) = 0
Here
tan 2x = 0 or tan 2x + 1 = 0
If tan 2x = 0
tan 2x = tan 0
We get
2x = nπ + 0, where n ∈ Z
x = nπ/2, where n ∈ Z
tan 2x + 1 = 0
We can write it as
tan 2x = – 1
So we get
Here
2x = nπ + 3π/4, where n ∈ Z
x = nπ/2 + 3π/8, where n ∈ Z
Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.
9. sin x + sin 3x + sin 5x = 0
Solution:
It is given that
sin x + sin 3x + sin 5x = 0
We can write it as
(sin x + sin 5x) + sin 3x = 0
Using the formula
By further calculation
2 sin 3x cos (-2x) + sin 3x = 0
It can be written as
2 sin 3x cos 2x + sin 3x = 0
By taking out the common terms
sin 3x (2 cos 2x + 1) = 0
Here
sin 3x = 0 or 2 cos 2x + 1 = 0
If sin 3x = 0
3x = nπ, where n ∈ Z
We get
x = nπ/3, where n ∈ Z
If 2 cos 2x + 1 = 0
cos 2x = – 1/2
By further simplification
= – cos π/3
= cos (π – π/3)
So we get
cos 2x = cos 2π/3
Here
Miscellaneous Exercise page: 81
Prove that:
1.
Solution:
We get
= 0
= RHS
2. (sin 3
x
+ sin
x
) sin
x
+ (cos 3
x
– cos
x
) cos
x
= 0
Solution:
Consider
LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
By further calculation
= sin 3x sin x + sin
2
x + cos 3x cos x – cos
2
x
Taking out the common terms
= cos 3x cos x + sin 3x sin x – (cos
2
x – sin
2
x)
Using the formula
cos (A – B) = cos A cos B + sin A sin B
= cos (3x – x) – cos 2x
So we get
= cos 2x – cos 2x
= 0
= RHS
3.
Solution:
Consider
LHS = (cos x + cos y)
2
+ (sin x – sin y)
2
By expanding using formula we get
= cos
2
x + cos
2
y + 2 cos x cos y + sin
2
x + sin
2
y – 2 sin x sin y
Grouping the terms
= (cos
2
x + sin
2
x) + (cos
2
y + sin
2
y) + 2 (cos x cos y – sin x sin y)
Using the formula cos (A + B) = (cos A cos B – sin A sin B)
= 1 + 1 + 2 cos (x + y)
By further calculation
= 2 + 2 cos (x + y)
Taking 2 as common
= 2 [1 + cos (x + y)]
From the formula cos 2A = 2 cos
2
A – 1
4.
Solution:
LHS = (cos x – cos y)
2
+ (sin x – sin y)
2
By expanding using formula
= cos
2
x + cos
2
y – 2 cos x cos y + sin
2
x + sin
2
y – 2 sin x sin y
Grouping the terms
= (cos
2
x + sin
2
x) + (cos
2
y + sin
2
y) – 2 (cos x cos y + sin x sin y)
Using the formula cos (A – B) = cos A cos B + sin A sin B
= 1 + 1 – 2 [cos (x – y)]
By further calculation
= 2 [1 – cos (x – y)]
From formula cos 2A = 1 – 2 sin
2
A
5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
Solution:
6.
Solution:
7.
Solution:
8. Find sin x/2, cos x/2 and tan x/2 in each of the following:
Solution:
cos x = -3/5
From the formula
9. cos x = -1/3, x in quadrant III
Solution:
10. sin x = 1/4, x in quadrant II
Solution:
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
This chapter has 6 exercises and a miscellaneous exercise to help students understand the concepts related to Trigonometric Functions clearly. BYJU’S provides all the concepts and solutions for Chapter 3 of Class 11 Maths with clear explanations and formulas. The PDF of Maths
NCERT Solutions for Class 11
Chapter 3 includes the topics and sub-topics listed below.
3.1 Introduction
The basic trigonometric ratios and identities are given here, along with the applications of trigonometric ratios in solving the word problems related to heights and distances.
3.2 Angles
3.2.1 Degree measure
3.2.2 Radian measure
3.2.3 Relation between radian and real numbers
3.2.4 Relation between degree and radian
In this section, different terms related to trigonometry are discussed, such as terminal side, initial sides, measuring an angle in degrees and radian, etc.
3.3 Trigonometric Functions
3.3.1 Sign of trigonometric functions
3.3.2 Domain and range of trigonometric functions
After studying this section, students are able to understand the generalised trigonometric functions with signs. Also, they can gain knowledge on domain and range of trigonometric functions with examples.
3.4 Trigonometric Functions of Sum and Difference of Two Angles
This section contains formulas related to the sum and difference of two angles in trigonometric functions.
Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
Studying the Trigonometric Functions of Class 11 enables the students to understand the following:
Introduction to Trigonometric Functions
Positive and negative angles
Measuring angles in radians and in degrees and conversion of one into other
Definition of trigonometric functions with the help of unit circle
Truth of the sin 2x + cos 2x = 1, for all x
Signs of trigonometric functions
Domain and range of trigonometric functions
Graphs of Trigonometric Functions
Expressing sin (x±y) and cos (x±y) in terms of sin x, sin y, cos x & cosy and their simple application, identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x
The general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a
Disclaimer –
Dropped Topics –
3.5 Trigonometric Equations (up to Exercise 3.4)
Last five points in the Summary
3.6 Proofs and Simple Applications of Sine and Cosine Formulae
Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 3
Q1
What are the topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths, as per the latest update of the CBSE Syllabus?
The topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths, as per the latest CBSE Syllabus (2023-24) are:
1. Introduction
2. Angles
3. Trigonometric Functions
4. Trigonometric Functions of sum and difference of two angles
Q2
How many exercises are there in Chapter 3 of NCERT Solutions for Class 11 Maths?
There are 4 exercises and one miscellaneous exercise in Chapter 3 of NCERT Solutions for Class 11 Maths. The number of questions in each exercise are mentioned below.
Exercise 3.1 – 7 Questions
Exercise 3.2 – 10 Questions
Exercise 3.3 – 25 Questions
Exercise 3.4 – 9 Questions
Miscellaneous Exercise – 10 Questions
Q3
What will I learn in Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths?
Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths covers the complex topics of trigonometric functions and their uses. This chapter consists of 4 sub-sections which deal with topics like measuring angles in radians and degrees and their interconversion. These concepts are explained in a detailed manner to improve logical thinking abilities among students. |
| Markdown | 
**NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions** are available at BYJU’S, which are prepared by our expert teachers. All these solutions are written as per the latest update on the CBSE Syllabus 2023-24 and its guidelines. BYJU’S provides step-by-step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3, Trigonometric Functions, comes under NCERT Class 11 Maths and is an important chapter for students. Though the chapter has numerous mathematical terms and formulae, BYJU’S has made [NCERT Solutions for Class 11 Maths](https://byjus.com/ncert-solutions-class-11-maths/) easy for the students to understand and remember with the usage of tricks.
Trigonometry has been developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is used in many areas, such as finding the heights of tides in the ocean, designing electronic circuits, etc. In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In **Class 11**, trigonometric ratios are generalised to trigonometric function and their properties. However, the [NCERT Solutions](https://byjus.com/ncert-solutions/) of BYJU’S help the students to attain more knowledge and score full marks in this chapter of the examination.
## [NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions]()
[Download PDF ]()
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In this section, a few important terms are defined, such as principal solution and general solution of trigonometric functions, which are explained using examples too.
[Exercise 3.1 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-1/) 7 Questions
[Exercise 3.2 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-2/) 10 Questions
[Exercise 3.3 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-3/) 25 Questions
[Exercise 3.4 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-4/) 9 Questions
[Miscellaneous Exercise On Chapter 3 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-miscellaneous-ex/) 10 Questions
### Access NCERT Solutions for Class 11 Maths Chapter 3
Exercise 3.1 page: 54
**1\. Find the radian measures corresponding to the following degree measures:**
**(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°**
**Solution:**
(iv) 520°
**2\. Find the degree measures corresponding to the following radian measures (Use π = 22/7)**
**(i) 11/16**
**(ii) -4**
**(iii) 5π/3**
**(iv) 7π/6**
**Solution:**
(i) 11/16
Here π radian = 180°
(ii) -4
Here π radian = 180°
(iii) 5π/3
Here π radian = 180°
We get
\= 300o
(iv) 7π/6
Here π radian = 180°
We get
\= 210o
**3\. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?**
**Solution:**
It is given that
No. of revolutions made by the wheel in
1 minute = 360
1 second = 360/60 = 6
We know that
The wheel turns an angle of 2π radian in one complete revolution.
In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian
Therefore, in one second, the wheel turns an angle of 12π radian.
**4\. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).**
**Solution:**
**5\. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.**
**Solution:**
The dimensions of the circle are
Diameter = 40 cm
Radius = 40/2 = 20 cm
Consider AB be as the chord of the circle i.e. length = 20 cm
In ΔOAB,
Radius of circle = OA = OB = 20 cm
Similarly AB = 20 cm
Hence, ΔOAB is an equilateral triangle.
θ = 60° = π/3 radian
In a circle of radius *r* unit, if an arc of length *l* unit subtends an angle *θ* radian at the centre
We get θ = 1/r
Therefore, the length of the minor arc of the chord is 20π/3 cm.
**6\. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.**
**Solution:**
**7\. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length**
**(i) 10 cm (ii) 15 cm (iii) 21 cm**
**Solution:**
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r
We know that r = 75 cm
(i) l = 10 cm
So we get
θ = 10/75 radian
By further simplification
θ = 2/15 radian
(ii) l = 15 cm
So we get
θ = 15/75 radian
By further simplification
θ = 1/5 radian
(iii) l = 21 cm
So we get
θ = 21/75 radian
By further simplification
θ = 7/25 radian
***
Exercise 3.2 page: 63
**Find the values of other five trigonometric functions in Exercises 1 to 5.**
**1\. cos x = -1/2, x lies in third quadrant.**
**Solution:**
**2\. sin x = 3/5, x lies in second quadrant.**
**Solution:**
It is given that
sin x = 3/5
We can write it as
We know that
sin2 x + cos2 x = 1
We can write it as
cos2 x = 1 – sin2 x
**3\. cot x = 3/4, x lies in third quadrant.**
**Solution:**
It is given that
cot x = 3/4
We can write it as
We know that
1 + tan2 x = sec2 x
We can write it as
1 + (4/3)2 = sec2 x
Substituting the values
1 + 16/9 = sec2 x
cos2 x = 25/9
sec x = ± 5/3
Here x lies in the third quadrant so the value of sec x will be negative
sec x = – 5/3
We can write it as
**4\. sec x = 13/5, x lies in fourth quadrant.**
**Solution:**
It is given that
sec x = 13/5
We can write it as
We know that
sin2 x + cos2 x = 1
We can write it as
sin2 x = 1 – cos2 x
Substituting the values
sin2 x = 1 – (5/13)2
sin2 x = 1 – 25/169 = 144/169
sin2 x = ± 12/13
Here x lies in the fourth quadrant so the value of sin x will be negative
sin x = – 12/13
We can write it as
**5\. tan x = -5/12, x lies in second quadrant.**
**Solution:**
It is given that
tan x = – 5/12
We can write it as
We know that
1 + tan2 x = sec2 x
We can write it as
1 + (-5/12)2 = sec2 x
Substituting the values
1 + 25/144 = sec2 x
sec2 x = 169/144
sec x = ± 13/12
Here x lies in the second quadrant so the value of sec x will be negative
sec x = – 13/12
We can write it as
**Find the values of the trigonometric functions in Exercises 6 to 10.**
**6\. sin 765°**
**Solution:**
We know that values of sin x repeat after an interval of 2π or 360°
So we get
By further calculation
\= sin 45o
\= 1/ **√** 2
**7\. cosec (–1410°)**
**Solution:**
We know that values of cosec x repeat after an interval of 2π or 360°
So we get
By further calculation
\= cosec 30o = 2
**8\. **
**Solution:**
We know that values of tan x repeat after an interval of π or 180°
So we get
By further calculation
We get
\= tan 60o
\= **√**3
**9\. **
**Solution:**
We know that values of sin x repeat after an interval of 2π or 360°
So we get
By further calculation
**10\. **
**Solution:**
We know that values of tan x repeat after an interval of π or 180°
So we get
By further calculation
***
Exercise 3.3 page: 73
**Prove that:**
**1\.**
****
**Solution:**
**2\.**
****
**Solution:**
Here
\= 1/2 + 4/4
\= 1/2 + 1
\= 3/2
\= RHS
**3\.**
****
**Solution:**
**4\.**
****
**Solution:**
**5\. Find the value of:**
**(i) sin 75o**
**(ii) tan 15o**
**Solution:**
(ii) tan 15°
It can be written as
\= tan (45° – 30°)
Using formula
**Prove the following:**
**6\.**
****
**Solution:**
**7\.**
****
**Solution:**
**8\.**
****
**Solution:**
**9\.**
****
**Solution:**
Consider
It can be written as
\= sin x cos x (tan x + cot x)
So we get
**10\. sin (*n* + 1)*x* sin (*n* + 2)*x* + cos (*n* + 1)*x* cos (*n* + 2)*x* = cos *x***
**Solution:**
LHS = sin (*n* + 1)*x* sin (*n* + 2)*x* + cos (*n* + 1)*x* cos (*n* + 2)*x*
**11\.**
****
**Solution:**
Consider
Using the formula
**12\. sin2 6*x* – sin2 4*x* = sin 2*x* sin 10*x***
**Solution:**
**13\. cos2 2*x* – cos2 6*x* = sin 4*x* sin 8*x***
**Solution:**
We get
\= \[2 cos 4x cos (-2x)\] \[-2 sin 4x sin (-2x)\]
It can be written as
\= \[2 cos 4*x* cos 2*x*\] \[–2 sin 4*x* (–sin 2*x*)\]
So we get
\= (2 sin 4*x* cos 4*x*) (2 sin 2*x* cos 2*x*)
\= sin 8x sin 4x
\= RHS
**14\. sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x**
**Solution:**
By further simplification
\= 2 sin 4x cos (– 2x) + 2 sin 4x
It can be written as
\= 2 sin 4x cos 2x + 2 sin 4x
Taking common terms
\= 2 sin 4x (cos 2x + 1)
Using the formula
\= 2 sin 4x (2 cos2 x – 1 + 1)
We get
\= 2 sin 4x (2 cos2 x)
\= 4cos2 x sin 4x
\= R.H.S.
**15\. cot 4*x* (sin 5*x* + sin 3*x*) = cot *x* (sin 5*x* – sin 3*x*)**
**Solution:**
Consider
LHS = cot 4x (sin 5x + sin 3x)
It can be written as
Using the formula
\= 2 cos 4x cos x
Hence, LHS = RHS.
**16\.**
****
**Solution:**
Consider
Using the formula
**17\.**
****
**Solution:**
**18\.**
****
**Solution:**
**19\.**
****
**Solution:**
**20\.**
****
**Solution:**
**21\.**
****
**Solution:**
**22\. cot *x* cot 2*x* – cot 2*x* cot 3*x* – cot 3*x* cot *x* = 1**
**Solution:**
**23\.**
****
**Solution:**
Consider
LHS = tan 4x = tan 2(2x)
By using the formula
**24\. cos 4*x* = 1 – 8sin2 *x* cos2 *x***
**Solution:**
Consider
LHS = cos 4x
We can write it as
\= cos 2(2*x*)
Using the formula cos 2*A* = 1 – 2 sin2 *A*
\= 1 – 2 sin2 2*x*
Again by using the formula sin2*A* = 2sin *A* cos *A*
\= 1 – 2(2 sin *x* cos *x*) 2
So we get
\= 1 – 8 sin2*x* cos2*x*
\= R.H.S.
**25\. cos 6*x* = 32 cos6 *x* – 48 cos4 *x* + 18 cos2 *x* – 1**
**Solution:**
Consider
L.H.S. = cos 6*x*
It can be written as
\= cos 3(2*x*)
Using the formula cos 3*A* = 4 cos3 *A* – 3 cos *A*
\= 4 cos3 2*x* – 3 cos 2*x*
Again by using formula cos 2*x* = 2 cos2 *x* – 1
\= 4 \[(2 cos2 *x* – 1)3 – 3 (2 cos2 *x* – 1)
By further simplification
\= 4 \[(2 cos2 *x*) 3 – (1)3 – 3 (2 cos2 *x*) 2 + 3 (2 cos2 *x*)\] – 6cos2 *x* + 3
We get
\= 4 \[8cos6*x* – 1 – 12 cos4*x* + 6 cos2*x*\] – 6 cos2*x* + 3
By multiplication
\= 32 cos6*x* – 4 – 48 cos4*x* + 24 cos2 *x* – 6 cos2*x* + 3
On further calculation
\= 32 cos6*x* – 48 cos4*x* + 18 cos2*x* – 1
\= R.H.S.
***
Exercise 3.4 PAGE: 78
**Find the principal and general solutions of the following equations:**
**1\. tan x = √3**
**Solution:**
**2\. sec x = 2**
**Solution:**
**3\. cot x = – √3**
**Solution:**
**4\. cosec x = – 2**
**Solution:**
**Find the general solution for each of the following equations:**
**5\. cos 4x = cos 2x**
**Solution:**
**6\. cos 3x + cos x – cos 2x = 0**
**Solution:**
**7\. sin 2x + cos x = 0**
**Solution:**
It is given that
sin 2x + cos x = 0
We can write it as
2 sin x cos x + cos x = 0
cos x (2 sin x + 1) = 0
cos x = 0 or 2 sin x + 1 = 0
Let cos x = 0
**8\. sec2 2x = 1 – tan 2x**
**Solution:**
It is given that
sec2 2x = 1 – tan 2x
We can write it as
1 + tan2 2x = 1 – tan 2x
tan2 2x + tan 2x = 0
Taking common terms
tan 2x (tan 2x + 1) = 0
Here
tan 2x = 0 or tan 2x + 1 = 0
If tan 2x = 0
tan 2x = tan 0
We get
2x = nπ + 0, where n ∈ Z
x = nπ/2, where n ∈ Z
tan 2x + 1 = 0
We can write it as
tan 2x = – 1
So we get
Here
2x = nπ + 3π/4, where n ∈ Z
x = nπ/2 + 3π/8, where n ∈ Z
Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.
**9\. sin x + sin 3x + sin 5x = 0**
**Solution:**
It is given that
sin x + sin 3x + sin 5x = 0
We can write it as
(sin x + sin 5x) + sin 3x = 0
Using the formula
By further calculation
2 sin 3x cos (-2x) + sin 3x = 0
It can be written as
2 sin 3x cos 2x + sin 3x = 0
By taking out the common terms
sin 3x (2 cos 2x + 1) = 0
Here
sin 3x = 0 or 2 cos 2x + 1 = 0
If sin 3x = 0
3x = nπ, where n ∈ Z
We get
x = nπ/3, where n ∈ Z
If 2 cos 2x + 1 = 0
cos 2x = – 1/2
By further simplification
\= – cos π/3
\= cos (π – π/3)
So we get
cos 2x = cos 2π/3
Here
***
Miscellaneous Exercise page: 81
**Prove that:**
**1\.**
****
**Solution:**
We get
\= 0
\= RHS
**2\. (sin 3*x* \+ sin *x*) sin *x* \+ (cos 3*x* – cos *x*) cos *x* \= 0**
**Solution:**
Consider
LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
By further calculation
\= sin 3x sin x + sin2 x + cos 3x cos x – cos2 x
Taking out the common terms
\= cos 3x cos x + sin 3x sin x – (cos2 x – sin2 x)
Using the formula
cos (A – B) = cos A cos B + sin A sin B
\= cos (3x – x) – cos 2x
So we get
\= cos 2x – cos 2x
\= 0
\= RHS
**3\.**
****
**Solution:**
Consider
LHS = (cos x + cos y) 2 + (sin x – sin y) 2
By expanding using formula we get
\= cos2 x + cos2 y + 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y
Grouping the terms
\= (cos2 x + sin2 x) + (cos2 y + sin2 y) + 2 (cos x cos y – sin x sin y)
Using the formula cos (A + B) = (cos A cos B – sin A sin B)
\= 1 + 1 + 2 cos (x + y)
By further calculation
\= 2 + 2 cos (x + y)
Taking 2 as common
\= 2 \[1 + cos (x + y)\]
From the formula cos 2A = 2 cos2 A – 1
**4\.**
****
**Solution:**
LHS = (cos x – cos y) 2 + (sin x – sin y) 2
By expanding using formula
\= cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y
Grouping the terms
\= (cos2 x + sin2 x) + (cos2 y + sin2 y) – 2 (cos x cos y + sin x sin y)
Using the formula cos (A – B) = cos A cos B + sin A sin B
\= 1 + 1 – 2 \[cos (x – y)\]
By further calculation
\= 2 \[1 – cos (x – y)\]
From formula cos 2A = 1 – 2 sin2 A
**5\. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x**
**Solution:**
**6\.**
****
**Solution:**
**7\.**
****
**Solution:**
**8\. Find sin x/2, cos x/2 and tan x/2 in each of the following:**
****
**Solution:**
cos x = -3/5
From the formula
**9\. cos x = -1/3, x in quadrant III**
**Solution:**
**10\. sin x = 1/4, x in quadrant II**
**Solution:**
***
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| *Also Access* |
| *[NCERT Exemplar for Class 11 Maths Chapter 3](https://byjus.com/ncert-exemplar-class-11-maths-chapter-3-trigonometric-functions/)* |
| *[CBSE Notes for Class 11 Maths Chapter 3](https://byjus.com/cbse-notes/cbse-class-11-maths-notes-chapter-3-trigonometric-functions/)* |
### NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
This chapter has 6 exercises and a miscellaneous exercise to help students understand the concepts related to Trigonometric Functions clearly. BYJU’S provides all the concepts and solutions for Chapter 3 of Class 11 Maths with clear explanations and formulas. The PDF of Maths [NCERT Solutions for Class 11](https://byjus.com/ncert-solutions-class-11/) Chapter 3 includes the topics and sub-topics listed below.
**3\.1 Introduction**
The basic trigonometric ratios and identities are given here, along with the applications of trigonometric ratios in solving the word problems related to heights and distances.
**3\.2 Angles**
3.2.1 Degree measure
3.2.2 Radian measure
3.2.3 Relation between radian and real numbers
3.2.4 Relation between degree and radian
In this section, different terms related to trigonometry are discussed, such as terminal side, initial sides, measuring an angle in degrees and radian, etc.
**3\.3 Trigonometric Functions**
3.3.1 Sign of trigonometric functions
3.3.2 Domain and range of trigonometric functions
After studying this section, students are able to understand the generalised trigonometric functions with signs. Also, they can gain knowledge on domain and range of trigonometric functions with examples.
**3\.4 Trigonometric Functions of Sum and Difference of Two Angles**
This section contains formulas related to the sum and difference of two angles in trigonometric functions.
#### Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
Studying the Trigonometric Functions of Class 11 enables the students to understand the following:
- Introduction to Trigonometric Functions
- Positive and negative angles
- Measuring angles in radians and in degrees and conversion of one into other
- Definition of trigonometric functions with the help of unit circle
- Truth of the sin 2x + cos 2x = 1, for all x
- Signs of trigonometric functions
- Domain and range of trigonometric functions
- Graphs of Trigonometric Functions
- Expressing sin (x±y) and cos (x±y) in terms of sin x, sin y, cos x & cosy and their simple application, identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x
- The general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a
**Disclaimer –**
**Dropped Topics –**
3\.5 Trigonometric Equations (up to Exercise 3.4)
Last five points in the Summary
3.6 Proofs and Simple Applications of Sine and Cosine Formulae
## Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 3
Q1
### What are the topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths, as per the latest update of the CBSE Syllabus?
The topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths, as per the latest CBSE Syllabus (2023-24) are:
1. Introduction
2. Angles
3. Trigonometric Functions
4. Trigonometric Functions of sum and difference of two angles
Q2
### How many exercises are there in Chapter 3 of NCERT Solutions for Class 11 Maths?
There are 4 exercises and one miscellaneous exercise in Chapter 3 of NCERT Solutions for Class 11 Maths. The number of questions in each exercise are mentioned below.
Exercise 3.1 – 7 Questions
Exercise 3.2 – 10 Questions
Exercise 3.3 – 25 Questions
Exercise 3.4 – 9 Questions
Miscellaneous Exercise – 10 Questions
Q3
### What will I learn in Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths?
Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths covers the complex topics of trigonometric functions and their uses. This chapter consists of 4 sub-sections which deal with topics like measuring angles in radians and degrees and their interconversion. These concepts are explained in a detailed manner to improve logical thinking abilities among students.
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| Readable Markdown | **NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions** are available at BYJU’S, which are prepared by our expert teachers. All these solutions are written as per the latest update on the CBSE Syllabus 2023-24 and its guidelines. BYJU’S provides step-by-step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3, Trigonometric Functions, comes under NCERT Class 11 Maths and is an important chapter for students. Though the chapter has numerous mathematical terms and formulae, BYJU’S has made [NCERT Solutions for Class 11 Maths](https://byjus.com/ncert-solutions-class-11-maths/) easy for the students to understand and remember with the usage of tricks.
Trigonometry has been developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is used in many areas, such as finding the heights of tides in the ocean, designing electronic circuits, etc. In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In **Class 11**, trigonometric ratios are generalised to trigonometric function and their properties. However, the [NCERT Solutions](https://byjus.com/ncert-solutions/) of BYJU’S help the students to attain more knowledge and score full marks in this chapter of the examination.
## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
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In this section, a few important terms are defined, such as principal solution and general solution of trigonometric functions, which are explained using examples too.
[Exercise 3.1 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-1/) 7 Questions
[Exercise 3.2 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-2/) 10 Questions
[Exercise 3.3 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-3/) 25 Questions
[Exercise 3.4 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-ex-3-4/) 9 Questions
[Miscellaneous Exercise On Chapter 3 Solutions](https://byjus.com/ncert-solutions-class-11-maths-chapter-3-trigonometric-functions-miscellaneous-ex/) 10 Questions
### Access NCERT Solutions for Class 11 Maths Chapter 3
Exercise 3.1 page: 54
**1\. Find the radian measures corresponding to the following degree measures:**
**(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°**
**Solution:**
(iv) 520°
**2\. Find the degree measures corresponding to the following radian measures (Use π = 22/7)**
**(i) 11/16**
**(ii) -4**
**(iii) 5π/3**
**(iv) 7π/6**
**Solution:**
(i) 11/16
Here π radian = 180°
(ii) -4
Here π radian = 180°
(iii) 5π/3
Here π radian = 180°
We get
\= 300o
(iv) 7π/6
Here π radian = 180°
We get
\= 210o
**3\. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?**
**Solution:**
It is given that
No. of revolutions made by the wheel in
1 minute = 360
1 second = 360/60 = 6
We know that
The wheel turns an angle of 2π radian in one complete revolution.
In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian
Therefore, in one second, the wheel turns an angle of 12π radian.
**4\. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).**
**Solution:**
**5\. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.**
**Solution:**
The dimensions of the circle are
Diameter = 40 cm
Radius = 40/2 = 20 cm
Consider AB be as the chord of the circle i.e. length = 20 cm
In ΔOAB,
Radius of circle = OA = OB = 20 cm
Similarly AB = 20 cm
Hence, ΔOAB is an equilateral triangle.
θ = 60° = π/3 radian
In a circle of radius *r* unit, if an arc of length *l* unit subtends an angle *θ* radian at the centre
We get θ = 1/r
Therefore, the length of the minor arc of the chord is 20π/3 cm.
**6\. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.**
**Solution:**
**7\. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length**
**(i) 10 cm (ii) 15 cm (iii) 21 cm**
**Solution:**
In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r
We know that r = 75 cm
(i) l = 10 cm
So we get
θ = 10/75 radian
By further simplification
θ = 2/15 radian
(ii) l = 15 cm
So we get
θ = 15/75 radian
By further simplification
θ = 1/5 radian
(iii) l = 21 cm
So we get
θ = 21/75 radian
By further simplification
θ = 7/25 radian
***
Exercise 3.2 page: 63
**Find the values of other five trigonometric functions in Exercises 1 to 5.**
**1\. cos x = -1/2, x lies in third quadrant.**
**Solution:**
**2\. sin x = 3/5, x lies in second quadrant.**
**Solution:**
It is given that
sin x = 3/5
We can write it as
We know that
sin2 x + cos2 x = 1
We can write it as
cos2 x = 1 – sin2 x
**3\. cot x = 3/4, x lies in third quadrant.**
**Solution:**
It is given that
cot x = 3/4
We can write it as
We know that
1 + tan2 x = sec2 x
We can write it as
1 + (4/3)2 = sec2 x
Substituting the values
1 + 16/9 = sec2 x
cos2 x = 25/9
sec x = ± 5/3
Here x lies in the third quadrant so the value of sec x will be negative
sec x = – 5/3
We can write it as
**4\. sec x = 13/5, x lies in fourth quadrant.**
**Solution:**
It is given that
sec x = 13/5
We can write it as
We know that
sin2 x + cos2 x = 1
We can write it as
sin2 x = 1 – cos2 x
Substituting the values
sin2 x = 1 – (5/13)2
sin2 x = 1 – 25/169 = 144/169
sin2 x = ± 12/13
Here x lies in the fourth quadrant so the value of sin x will be negative
sin x = – 12/13
We can write it as
**5\. tan x = -5/12, x lies in second quadrant.**
**Solution:**
It is given that
tan x = – 5/12
We can write it as
We know that
1 + tan2 x = sec2 x
We can write it as
1 + (-5/12)2 = sec2 x
Substituting the values
1 + 25/144 = sec2 x
sec2 x = 169/144
sec x = ± 13/12
Here x lies in the second quadrant so the value of sec x will be negative
sec x = – 13/12
We can write it as
**Find the values of the trigonometric functions in Exercises 6 to 10.**
**6\. sin 765°**
**Solution:**
We know that values of sin x repeat after an interval of 2π or 360°
So we get
By further calculation
\= sin 45o
\= 1/ **√** 2
**7\. cosec (–1410°)**
**Solution:**
We know that values of cosec x repeat after an interval of 2π or 360°
So we get
By further calculation
\= cosec 30o = 2
**8\. **
**Solution:**
We know that values of tan x repeat after an interval of π or 180°
So we get
By further calculation
We get
\= tan 60o
\= **√**3
**9\. **
**Solution:**
We know that values of sin x repeat after an interval of 2π or 360°
So we get
By further calculation
**10\. **
**Solution:**
We know that values of tan x repeat after an interval of π or 180°
So we get
By further calculation
***
Exercise 3.3 page: 73
**Prove that:**
**1\.**
****
**Solution:**
**2\.**
****
**Solution:**
Here
\= 1/2 + 4/4
\= 1/2 + 1
\= 3/2
\= RHS
**3\.**
****
**Solution:**
**4\.**
****
**Solution:**
**5\. Find the value of:**
**(i) sin 75o**
**(ii) tan 15o**
**Solution:**
(ii) tan 15°
It can be written as
\= tan (45° – 30°)
Using formula
**Prove the following:**
**6\.**
****
**Solution:**
**7\.**
****
**Solution:**
**8\.**
****
**Solution:**
**9\.**
****
**Solution:**
Consider
It can be written as
\= sin x cos x (tan x + cot x)
So we get
**10\. sin (*n* + 1)*x* sin (*n* + 2)*x* + cos (*n* + 1)*x* cos (*n* + 2)*x* = cos *x***
**Solution:**
LHS = sin (*n* + 1)*x* sin (*n* + 2)*x* + cos (*n* + 1)*x* cos (*n* + 2)*x*
**11\.**
****
**Solution:**
Consider
Using the formula
**12\. sin2 6*x* – sin2 4*x* = sin 2*x* sin 10*x***
**Solution:**
**13\. cos2 2*x* – cos2 6*x* = sin 4*x* sin 8*x***
**Solution:**
We get
\= \[2 cos 4x cos (-2x)\] \[-2 sin 4x sin (-2x)\]
It can be written as
\= \[2 cos 4*x* cos 2*x*\] \[–2 sin 4*x* (–sin 2*x*)\]
So we get
\= (2 sin 4*x* cos 4*x*) (2 sin 2*x* cos 2*x*)
\= sin 8x sin 4x
\= RHS
**14\. sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x**
**Solution:**
By further simplification
\= 2 sin 4x cos (– 2x) + 2 sin 4x
It can be written as
\= 2 sin 4x cos 2x + 2 sin 4x
Taking common terms
\= 2 sin 4x (cos 2x + 1)
Using the formula
\= 2 sin 4x (2 cos2 x – 1 + 1)
We get
\= 2 sin 4x (2 cos2 x)
\= 4cos2 x sin 4x
\= R.H.S.
**15\. cot 4*x* (sin 5*x* + sin 3*x*) = cot *x* (sin 5*x* – sin 3*x*)**
**Solution:**
Consider
LHS = cot 4x (sin 5x + sin 3x)
It can be written as
Using the formula
\= 2 cos 4x cos x
Hence, LHS = RHS.
**16\.**
****
**Solution:**
Consider
Using the formula
**17\.**
****
**Solution:**
**18\.**
****
**Solution:**
**19\.**
****
**Solution:**
**20\.**
****
**Solution:**
**21\.**
****
**Solution:**
**22\. cot *x* cot 2*x* – cot 2*x* cot 3*x* – cot 3*x* cot *x* = 1**
**Solution:**
**23\.**
****
**Solution:**
Consider
LHS = tan 4x = tan 2(2x)
By using the formula
**24\. cos 4*x* = 1 – 8sin2 *x* cos2 *x***
**Solution:**
Consider
LHS = cos 4x
We can write it as
\= cos 2(2*x*)
Using the formula cos 2*A* = 1 – 2 sin2 *A*
\= 1 – 2 sin2 2*x*
Again by using the formula sin2*A* = 2sin *A* cos *A*
\= 1 – 2(2 sin *x* cos *x*) 2
So we get
\= 1 – 8 sin2*x* cos2*x*
\= R.H.S.
**25\. cos 6*x* = 32 cos6 *x* – 48 cos4 *x* + 18 cos2 *x* – 1**
**Solution:**
Consider
L.H.S. = cos 6*x*
It can be written as
\= cos 3(2*x*)
Using the formula cos 3*A* = 4 cos3 *A* – 3 cos *A*
\= 4 cos3 2*x* – 3 cos 2*x*
Again by using formula cos 2*x* = 2 cos2 *x* – 1
\= 4 \[(2 cos2 *x* – 1)3 – 3 (2 cos2 *x* – 1)
By further simplification
\= 4 \[(2 cos2 *x*) 3 – (1)3 – 3 (2 cos2 *x*) 2 + 3 (2 cos2 *x*)\] – 6cos2 *x* + 3
We get
\= 4 \[8cos6*x* – 1 – 12 cos4*x* + 6 cos2*x*\] – 6 cos2*x* + 3
By multiplication
\= 32 cos6*x* – 4 – 48 cos4*x* + 24 cos2 *x* – 6 cos2*x* + 3
On further calculation
\= 32 cos6*x* – 48 cos4*x* + 18 cos2*x* – 1
\= R.H.S.
***
Exercise 3.4 PAGE: 78
**Find the principal and general solutions of the following equations:**
**1\. tan x = √3**
**Solution:**
**2\. sec x = 2**
**Solution:**
**3\. cot x = – √3**
**Solution:**
**4\. cosec x = – 2**
**Solution:**
**Find the general solution for each of the following equations:**
**5\. cos 4x = cos 2x**
**Solution:**
**6\. cos 3x + cos x – cos 2x = 0**
**Solution:**
**7\. sin 2x + cos x = 0**
**Solution:**
It is given that
sin 2x + cos x = 0
We can write it as
2 sin x cos x + cos x = 0
cos x (2 sin x + 1) = 0
cos x = 0 or 2 sin x + 1 = 0
Let cos x = 0
**8\. sec2 2x = 1 – tan 2x**
**Solution:**
It is given that
sec2 2x = 1 – tan 2x
We can write it as
1 + tan2 2x = 1 – tan 2x
tan2 2x + tan 2x = 0
Taking common terms
tan 2x (tan 2x + 1) = 0
Here
tan 2x = 0 or tan 2x + 1 = 0
If tan 2x = 0
tan 2x = tan 0
We get
2x = nπ + 0, where n ∈ Z
x = nπ/2, where n ∈ Z
tan 2x + 1 = 0
We can write it as
tan 2x = – 1
So we get
Here
2x = nπ + 3π/4, where n ∈ Z
x = nπ/2 + 3π/8, where n ∈ Z
Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.
**9\. sin x + sin 3x + sin 5x = 0**
**Solution:**
It is given that
sin x + sin 3x + sin 5x = 0
We can write it as
(sin x + sin 5x) + sin 3x = 0
Using the formula
By further calculation
2 sin 3x cos (-2x) + sin 3x = 0
It can be written as
2 sin 3x cos 2x + sin 3x = 0
By taking out the common terms
sin 3x (2 cos 2x + 1) = 0
Here
sin 3x = 0 or 2 cos 2x + 1 = 0
If sin 3x = 0
3x = nπ, where n ∈ Z
We get
x = nπ/3, where n ∈ Z
If 2 cos 2x + 1 = 0
cos 2x = – 1/2
By further simplification
\= – cos π/3
\= cos (π – π/3)
So we get
cos 2x = cos 2π/3
Here
***
Miscellaneous Exercise page: 81
**Prove that:**
**1\.**
****
**Solution:**
We get
\= 0
\= RHS
**2\. (sin 3*x* \+ sin *x*) sin *x* \+ (cos 3*x* – cos *x*) cos *x* \= 0**
**Solution:**
Consider
LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
By further calculation
\= sin 3x sin x + sin2 x + cos 3x cos x – cos2 x
Taking out the common terms
\= cos 3x cos x + sin 3x sin x – (cos2 x – sin2 x)
Using the formula
cos (A – B) = cos A cos B + sin A sin B
\= cos (3x – x) – cos 2x
So we get
\= cos 2x – cos 2x
\= 0
\= RHS
**3\.**
****
**Solution:**
Consider
LHS = (cos x + cos y) 2 + (sin x – sin y) 2
By expanding using formula we get
\= cos2 x + cos2 y + 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y
Grouping the terms
\= (cos2 x + sin2 x) + (cos2 y + sin2 y) + 2 (cos x cos y – sin x sin y)
Using the formula cos (A + B) = (cos A cos B – sin A sin B)
\= 1 + 1 + 2 cos (x + y)
By further calculation
\= 2 + 2 cos (x + y)
Taking 2 as common
\= 2 \[1 + cos (x + y)\]
From the formula cos 2A = 2 cos2 A – 1
**4\.**
****
**Solution:**
LHS = (cos x – cos y) 2 + (sin x – sin y) 2
By expanding using formula
\= cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y
Grouping the terms
\= (cos2 x + sin2 x) + (cos2 y + sin2 y) – 2 (cos x cos y + sin x sin y)
Using the formula cos (A – B) = cos A cos B + sin A sin B
\= 1 + 1 – 2 \[cos (x – y)\]
By further calculation
\= 2 \[1 – cos (x – y)\]
From formula cos 2A = 1 – 2 sin2 A
**5\. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x**
**Solution:**
**6\.**
****
**Solution:**
**7\.**
****
**Solution:**
**8\. Find sin x/2, cos x/2 and tan x/2 in each of the following:**
****
**Solution:**
cos x = -3/5
From the formula
**9\. cos x = -1/3, x in quadrant III**
**Solution:**
**10\. sin x = 1/4, x in quadrant II**
**Solution:**
***
### NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
This chapter has 6 exercises and a miscellaneous exercise to help students understand the concepts related to Trigonometric Functions clearly. BYJU’S provides all the concepts and solutions for Chapter 3 of Class 11 Maths with clear explanations and formulas. The PDF of Maths [NCERT Solutions for Class 11](https://byjus.com/ncert-solutions-class-11/) Chapter 3 includes the topics and sub-topics listed below.
**3\.1 Introduction**
The basic trigonometric ratios and identities are given here, along with the applications of trigonometric ratios in solving the word problems related to heights and distances.
**3\.2 Angles**
3.2.1 Degree measure
3.2.2 Radian measure
3.2.3 Relation between radian and real numbers
3.2.4 Relation between degree and radian
In this section, different terms related to trigonometry are discussed, such as terminal side, initial sides, measuring an angle in degrees and radian, etc.
**3\.3 Trigonometric Functions**
3.3.1 Sign of trigonometric functions
3.3.2 Domain and range of trigonometric functions
After studying this section, students are able to understand the generalised trigonometric functions with signs. Also, they can gain knowledge on domain and range of trigonometric functions with examples.
**3\.4 Trigonometric Functions of Sum and Difference of Two Angles**
This section contains formulas related to the sum and difference of two angles in trigonometric functions.
#### Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
Studying the Trigonometric Functions of Class 11 enables the students to understand the following:
- Introduction to Trigonometric Functions
- Positive and negative angles
- Measuring angles in radians and in degrees and conversion of one into other
- Definition of trigonometric functions with the help of unit circle
- Truth of the sin 2x + cos 2x = 1, for all x
- Signs of trigonometric functions
- Domain and range of trigonometric functions
- Graphs of Trigonometric Functions
- Expressing sin (x±y) and cos (x±y) in terms of sin x, sin y, cos x & cosy and their simple application, identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x
- The general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a
**Disclaimer –**
**Dropped Topics –**
3\.5 Trigonometric Equations (up to Exercise 3.4)
Last five points in the Summary
3.6 Proofs and Simple Applications of Sine and Cosine Formulae
## Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 3
Q1
### What are the topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths, as per the latest update of the CBSE Syllabus?
The topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths, as per the latest CBSE Syllabus (2023-24) are:
1. Introduction
2. Angles
3. Trigonometric Functions
4. Trigonometric Functions of sum and difference of two angles
Q2
### How many exercises are there in Chapter 3 of NCERT Solutions for Class 11 Maths?
There are 4 exercises and one miscellaneous exercise in Chapter 3 of NCERT Solutions for Class 11 Maths. The number of questions in each exercise are mentioned below.
Exercise 3.1 – 7 Questions
Exercise 3.2 – 10 Questions
Exercise 3.3 – 25 Questions
Exercise 3.4 – 9 Questions
Miscellaneous Exercise – 10 Questions
Q3
### What will I learn in Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths?
Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths covers the complex topics of trigonometric functions and their uses. This chapter consists of 4 sub-sections which deal with topics like measuring angles in radians and degrees and their interconversion. These concepts are explained in a detailed manner to improve logical thinking abilities among students. |
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