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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are provided here. Trigonometry class 11 has various real-time applications. It is used to solve height and distance problems. This chapter gives an introduction to basic properties and identities of trigonometric functions and questions based on the same are answered in class 11 trigonometric functions NCERT solutions. Students can practice NCERT Solutions that are prepared by Careers360 expert team according to CBSE latest syllabus 2023. The chapter 3 maths class 11 trigonometric identities are used in multiple other chapters of mathematics and throughout the NCERT physics for both class 11 and 12 also. So it is very important to memories and understand the basic identities and properties of trigonometric functions. Trigonometric functions class 11 NCERT solutions will help you for the same.
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
Suggested: JEE Main: high scoring chapters | Past 10 year's papers
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Command on this NCERT syllabus is required to understand the concepts of inverse trigonometric functions which will be taught in class 12. You should practice more in order to get clarity of the concepts. First, try to solve NCERT textbook exercise problems. If you getting difficulties in doing so, you can take help from solutions of NCERT for class 11 maths chapter 3 trigonometric functions. Check all NCERT solutions at a single place which will be helpful when you are not able to solve the NCERT questions. Here you will get NCERT solutions for class 11 also.
Also Read
Angle Conversion:
Radian Measure = π/180 × Degree Measure
Degree Measure = 180/π × Radian Measure
Trigonometric Ratios:
sin θ = P / H
cos θ = B / H
tan θ = P / B
cosec θ = H / P
sec θ = H / B
cot θ = B / P
Reciprocal Trigonometric Ratios:
sin θ = 1 / cosec θ
cosec θ = 1 / sin θ
cos θ = 1 / sec θ
sec θ = 1 / cos θ
tan θ = 1 / cot θ
cot θ = 1 / tan θ
Trigonometric Ratios of Complementary Angles:
sin (90° – θ) = cos θ
cos (90° – θ) = sin θ
tan (90° – θ) = cot θ
cot (90° – θ) = tan θ
sec (90° – θ) = cosec θ
cosec (90° – θ) = sec θ
Periodic Trigonometric Ratios:
sin(π/2-θ) = cos θ
cos(π/2-θ) = sin θ
sin(π-θ) = sin θ
cos(π-θ) = -cos θ
sin(π+θ) = -sin θ
cos(π+θ) = -cos θ
sin(2π-θ) = -sin θ
cos(2π-θ) = cos θ
Trigonometric Identities:
sin^{²} θ + cos^{²} θ = 1
cosec^{²} θ – cot^{²} θ = 1
sec^{²} θ – tan^{²} θ = 1
Product to Sum Formulas:
sin x sin y = 1/2 [cos(x–y) − cos(x+y)]
cos x cos y = 1/2[cos(x–y) + cos(x+y)]
sin x cos y = 1/2[sin(x+y) + sin(x−y)]
cos x sin y = 1/2[sin(x+y) – sin(x−y)]
Sum to Product Formulas:
sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]
sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]
cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]
cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]
General Trigonometric Formulas:
sin (x+y) = sin x × cos y + cos x × sin y
cos(x+y) = cosx × cosy − sinx × siny
cos(x–y) = cosx × cosy + sinx × siny
sin(x–y) = sinx × cosy − cosx × siny
Sum and Difference Formulas for tan:
tan (x+y) = (tan x + tan y) / (1 − tan x tan y)
tan (x−y) = (tan x − tan y) / (1 + tan x tan y)
Double Angle Formulas for tan:
tan 2θ = (2 tan θ) / (1 - tan²θ)
Triple Angle Formulas for sin, cos, and tan:
sin 3θ = 3sin θ – 4sin^{³}θ
cos 3θ = 4cos^{³}θ – 3cos θ
tan 3θ = [3tan θ – tan^{³}θ] / [1 − 3tan^{²}θ]
Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions for CBSE Exam.
NCERT trigonometric functions class 11 questions and answers - Exercise: 3.1
Question:1 Find the radian measures corresponding to the following degree measures:
(i)
(ii)
(iii)
(iv)
Answer:
It is solved using relation between degree and radian
(i)
We know that = radian
So, radian
radian radian
(ii)
We know that
Now, we know that radian
So, radian radian
(iii)
We know that
radian
So, radian
(iv)
We know that
radian
So, radian radian
Question:2 Find the degree measures corresponding to the following radian measures. (Use )
Answer:
(1)
We know that
radian
So, (we need to take )
(we use and 1' = 60'')
Here 1' represents 1 minute and 60" represents 60 seconds
Now,
(ii) -4
We know that
radian (we need to take )
So, -4 radian =
(we use and 1' = 60'')
(iii)
We know that
radian (we need to take )
So,
(iv)
We know that
radian (we need to take )
So,
Question:3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Answer:
Number of revolutions made by the wheel in 1 minute = 360
Number of revolutions made by the wheel in 1 second =
( 1 minute = 60 seconds)
In one revolutions wheel will cover radian
So, in 6 revolutions it will cover = radian
In 1 the second wheel will turn radian
Answer:
We know that
( where l is the length of the arc, r is the radius of the circle and is the angle subtended)
here r = 100 cm
and l = 22 cm
Now,
We know that
So,
Angle subtended at the centre of a circle
Answer:
Given :- radius (r)of circle =
length of chord = 20 cm
We know that
(r = 20cm , l = ? , = ?)
Now,
AB is the chord of length 20cm and OA and OB are radii of circle 20 cm each
The angle subtended by OA and OB at centre =
OA = OB = AB
OAB is equilateral triangle
So, each angle equilateral is
Now, we have and r
So,
the length of the minor arc of the chord (l) = cm
Answer:
Given:-
and
We need to find the ratio of their radii
We know that arc length
So,
Now,
( )
So,
is the ratio of their radii
Answer:
(i) We know that
Now,
r = 75cm
l = 10cm
So,
(ii) We know that
Now,
r = 75cm
l = 15cm
So,
(iii) We know that
Now,
r = 75cm
l = 21cm
So,
NCERT class 11 maths chapter 3 question answer - Exercise: 3.2
Question:1 Find the values of other five trigonometric functions , x lies in third quadrant.
Answer:
Solution
x lies in III quadrants. Therefore sec x is negative
x lies in III quadrants. Therefore sin x is negative
x lies in III quadrants. Therefore cosec x is negative
x lies in III quadrants. Therefore tan x is positive
x lies in III quadrants. Therefore cot x is positive
Question:2 Find the values of other five trigonometric functions x lies in second quadrant.
Answer:
Solution
x lies in the second quadrant. Therefore cosec x is positive
x lies in the second quadrant. Therefore cos x is negative
x lies in the second quadrant. Therefore sec x is negative
x lies in the second quadrant. Therefore tan x is negative
x lies in the second quadrant. Therefore cot x is negative
Question:3 Find the values of other five trigonometric functions , x lies in third quadrant.
Answer:
Solution
x lies in x lies in third quadrant. therefore sec x is negative
x lies in x lies in third quadrant. Therefore sin x is negative
Question:4 Find the values of other five trigonometric functions , x lies in fourth quadrant.
Answer:
Solution
lies in fourth quadrant. Therefore sin x is negative
Question:5 Find the values of the other five trigonometric functions , x lies in second quadrant.
Answer:
x lies in second quadrant. Therefore the value of sec x is negative
x lies in the second quadrant. Therefore the value of sin x is positive
Question:6 Find the values of the trigonometric functions
Answer:
We know that values of sin x repeat after an interval of
Question:7 Find the values of the trigonometric functions
Answer:
We know that value of cosec x repeats after an interval of
or
Question:8 Find the values of the trigonometric functions
Answer:
We know that tan x repeats after an interval of or 180 degree
Question:9 Find the values of the trigonometric functions
Answer:
We know that sin x repeats after an interval of
Question:10 Find the values of the trigonometric functions
Answer:
We know that cot x repeats after an interval of
NCERT class 11 maths chapter 3 question answer - Exercise: 3.3
Question:1 Prove that
Answer:
We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:
= R.H.S.
Question:3 Prove that
Answer:
We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)
R.H.S.
Question:5(i) Find the value of
Answer:
We know that
(sin(x+y)=sinxcosy + cosxsiny)
Using this idendity
Question:6 Prove the following:
Answer:
Multiply and divide by 2 both cos and sin functions
We get,
Now, we know that
2cosAcosB = cos(A+B) + cos(A-B) -(i)
-2sinAsinB = cos(A+B) - cos(A-B) -(ii)
We use these two identities
In our question A =
B =
So,
As we know that
By using this
R.H.S
Question:8 Prove the following
Answer:
As we know that,
, ,
and
By using these our equation simplify to
R.H.S.
Question:9 Prove the following
Answer:
We know that
So, by using these our equation simplifies to
R.H.S.
Question:10 Prove the following
Answer:
Multiply and divide by 2
Now by using identities
-2sinAsinB = cos(A+B) - cos(A-B)
2cosAcosB = cos(A+B) + cos(A-B)
R.H.S.
Question:11 Prove the following
Answer:
We know that
[ cos(A+B) - cos (A-B) = -2sinAsinB ]
By using this identity
R.H.S.
Question:12 Prove the following
Answer:
We know that
So,
Now, we know that
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx
Now,
2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)
by using these identities
2cos5x sin5x = sin10x - 0
2sinx cosx = sin2x + 0
hence
Question:13 Prove the following
Answer:
As we know that
Now
By using these identities
cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( sin(-x) = -sin x
cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x
So our equation becomes
R.H.S.
Question:14 Prove the following
Answer:
We know that
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x
sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( cos(-x) = cos x)
So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( )
=2sin4x( +1 )
=2sin4x( )
=
R.H.S.
Question:15 Prove the following
Answer:
We know that
By using this , we get
sin5x + sin3x = 2sin4xcosx
now nultiply and divide by sin x
Now we know that
By using this our equation becomes
R.H.S.
Question:22 prove the following
Answer:
cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)
and we know that
So,
= cotx cot2x - (cot2xcotx -1)
= cotx cot2x - cot2xcotx +1
= 1 = R.H.S.
Question:23 Prove that
Answer:
We know that
and we can write tan 4x = tan 2(2x)
So, =
=
=
= = R.H.S.
Question:24 Prove the following
Answer:
We know that
We use this in our problem
cos 4x = cos 2(2x)
=
=
= = R.H.S.
Question:25 Prove the following
Answer:
We know that
cos 3x = 4 - 3cos x
we use this in our problem
we can write cos 6x as cos 3(2x)
cos 3(2x) = 4 - 3 cos 2x
= -
=
= 32 - 4 - 48 + 24 -
= 32 - 48 + 18 - 1 = R.H.S.
NCERT class 11 maths chapter 3 question answer - Exercise: 3.4
Question:1 Find the principal and general solutions of the following equations:
Answer:
It is given that given
Now, we know that and
Therefore,
the principal solutions of the equation are
Now,
The general solution is
where and Z denotes sets of integer
Therefore, the general solution of the equation is where and Z denotes sets of integer
Question:2 Find the principal and general solutions of the following equations:
Answer:
We know that value of and
Therefore the principal solutions are x =
We know that value of sec x repeats after an interval of
So, by this we can say that
the general solution is x = where n Z
Question:3 Find the principal and general solutions of the following equations:
Answer:
we know that and we know that
Similarly , the value for
Therefore, principal solution is x =
We also know that the value of cot x repeats after an interval of
There the general solution is x =
Question:4 Find the principal and general solutions of the following equations:
Answer:
We know that
and also
So,
and
So, the principal solutions are
Now,
Therefore, the general solution is
where
Question:5 Find the general solution for each of the following equation
Answer:
cos4x = cos2x
cos4x - cos2x = 0
We know that
We use this identity
cos 4x - cos 2x = -2sin3xsinx
-2sin3xsinx = 0 sin3xsinx=0
So, by this we can that either
sin3x = 0 or sinx = 0
3x = x =
x = x =
Therefore, the general solution is
Question:6 Find the general solution of the following equation
Answer:
We know that
We use these identities
(cos3x + cosx) - cos2x = 2cos2xcosx -cos2x = 0
= cos2x(2cosx-1) = 0
So, either
cos2x = 0 or
the general solution is
Question:7 Find the general solution of the following equation
Answer:
sin2x + cosx = 0
We know that
sin2x = 2sinxcosx
So,
2sinxcosx + cosx = 0
cosx(2sinx + 1) = 0
So, we can say that either
cosx = 0 or 2sinx + 1 = 0
Therefore, the general solution is
Question:8 Find the general solution of the following equation
Answer:
We know that
So,
either
tan2x = 0 or tan2x = -1 ( )
2x =
Where n Z
Question:9 Find the general solution of the following equation
Answer:
We know that
We use this identity in our problem
Now our problem simplifeis to
= 0
take sin3x common
So, either
sin3x = 0 or
Where
Class 11 maths chapter 3 NCERT solutions - Miscellaneous Exercise
Question:1 Prove that
Answer:
We know that
cos A+ cos B =
we use this in our problem
( we know that cos(-x) = cos x )
again use the above identity
we know that
= 0
So,
= 0 = R.H.S.
Question:2 Prove that
Answer:
We know that
and
We use this in our problem
= +
= (4sinx - 4 )sinx + (4 - 4cos x)cosx
now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4 (1 - ) - 4 (1 - )
= 4 - 4
= 0 = R.H.S.
Question:3 Prove that
Answer:
We know that
and
We use these two in our problem
and
= +
= 1 + 2cosxcosy + 1 - 2sinxsiny
= 2 + 2(cosxcosy - sinxsiny)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
=
= R.H.S.
Question:4 Prove that
Answer:
We know that
and
We use these two in our problem
and
= +
= 1 - 2cosxcosy + 1 - 2sinxsiny
= 2 - 2(cosxcosy + sinxsiny)
= 2 - 2cos(x - y)
= 2(1 - cos(x - y) )
Now we can write
so
= R.H.S.
Question:5 Prove that
Answer:
we know that
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and sin5x tp get sin4x
take 2sin4x common
= 2sin4x(cos3x + cosx)
Now,
We know that
We use this
=
= 2sin4x( )
= 4cosxcos2xsin4x = R.H.S.
Question:6 Prove that
Answer:
We know that
and
We use these two identities in our problem
sin7x + sin5x = =
sin 9x + sin 3x = =
cos 7x + cos5x = =
cos 9x + cos3x = =
=
= = R.H.S.
Question:7 Prove that
Answer:
We know that
we use these identities
sin2x + = 2sinx cosx +
take 2 sinx common
= R.H.S.
Question:8 Find in , x in quadrant II
Answer:
tan x =
We know that ,
=
=
x lies in II quadrant thats why sec x is -ve
So,
Now, =
We know that,
( )
= =
=
=
= =
x lies in II quadrant so value of is +ve
=
we know that
= 1 - =
x lies in II quadrant So value of sin x is +ve
Question:9 Find in , x in quadrant III
Answer:
We know that
cos x =
cos x + 1
= + 1 = =
Now,
we know that
cos x =
= 1 - = =
Because is +ve in given quadrant
Question:10 Find in ,x in quadrant II
Answer:
all functions are positive in this range
We know that
= 1 - = =
cos x = (cos x is -ve in II quadrant)
We know that
cosx =
(because all functions are posititve in given range)
similarly,
cos x =
(because all functions are posititve in given range)
Interested students can study Class 11 NCERT Solutions at a single place. Here are some highlights of Maths NCERT Solutions for Class 11 Chapter 3 encompasses the following topics and sub-topics:
Interested students can study Trigonometric Functions Exercise using following links-
chapter-1 | Sets |
chapter-2 | Relations and Functions |
chapter-3 | Trigonometric Functions |
chapter-4 | Principle of Mathematical Induction |
chapter-5 | Complex Numbers and Quadratic equations |
chapter-6 | Linear Inequalities |
chapter-7 | Permutation and Combinations |
chapter-8 | Binomial Theorem |
chapter-9 | Sequences and Series |
chapter-10 | Straight Lines |
chapter-11 | Conic Section |
chapter-12 | Three Dimensional Geometry |
chapter-13 | Limits and Derivatives |
chapter-14 | Mathematical Reasoning |
chapter-15 | Statistics |
chapter-16 | Probability |
Comprehensive Coverage: The solutions provide a comprehensive explanation of all the topics covered in Chapter 3, ensuring a strong understanding of trigonometric functions.
Step-by-Step Solutions: Each problem and example in the NCERT textbook is solved step by step, making it easier for students to follow the logical progression of trigonometric concepts.
Clear and Concise Language: The solutions are presented in clear and concise language, making complex trigonometric concepts more accessible to students.
Trigonometric Functions Basic Identities
The above identities you may have studied in your high school classes also. Here are a few more identities that you have to remember and understand from the NCERT class 11 maths chapter 3 trigonometric functions
Some conditional identities from the NCERT solutions for class 11 maths chapter 3 Trigonometric Functions
If angles x, y and (x ± y) is not an odd multiple of π 2, then
If angles x, y and (x ± y) is not a multiple of π, then
There are a few more identities used in the NCERT solutions for class 11 maths chapter 3 trigonometric functions which can be derived using the above identities. Try to derive it by your self.
Happy Reading !!!
In the chapter 3 class 11 maths Trigonometric functions, identities of trigonometric functions, trigonometric functions of some standard angles, trigonometric functions of sum and difference of two angles and trigonometric equations are the important topics of this chapter.
NCERT solutions will help the students, if the stuck while solving the NCERT problems. Trigonometry class 11 solutions provides practice problems for students that help them to command and get indepth understanding. finally class 11 maths trigonometry help to build strong foundation for higher classes. For ease Interested students can study trigonometric functions class 11 pdf both online and offline. therefore students can practice class 11 maths chapter 3 solutions.
Complex Numbers and Quadratic Equations : This chapter introduces students to complex numbers, which can be a new and abstract concept for many. Understanding operations with complex numbers and solving quadratic equations involving complex roots can be challenging.
Limits and Derivatives : Calculus, in general, tends to be difficult for many students, especially when they encounter the concept of limits for the first time. Understanding the concept of limits and their applications, as well as learning differentiation rules and techniques, can be challenging.
Permutations and Combinations : This chapter involves various counting principles and problem-solving techniques, which can be confusing for some students. Understanding when to use permutations versus combinations and solving problems involving both can be tricky.
Binomial Theorem : The binomial theorem involves expanding expressions raised to positive integer powers, which can require a good understanding of algebraic manipulation and pattern recognition.
Statistics : This chapter involves understanding various measures of central tendency and dispersion, as well as probability distributions. It can be challenging for some students to grasp the concepts and apply them to solve problems.
Trigonometric Functions : Trigonometry, particularly when it involves proving identities and solving trigonometric equations, can be challenging due to the abstract nature of the concepts and the numerous identities to remember.
Ultimately, the toughest chapter can vary based on individual preferences and strengths. Some students may find certain chapters easier than others based on their prior knowledge or interests in specific topics. Regular practice, seeking clarification when needed, and utilizing resources such as textbooks, online tutorials, and practice problems can help students overcome challenges in any chapter.
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No, CBSE doesn’t provided NCERT solutions for any class or subject.
Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.
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