# NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions: Trigonometry has various real-time applications. It is used to solve height and distance problems. This cahpter gives an introduction to basic properties and identities of trigonometric functions and questions based on the same are answered in NCERT solutions for class 11 maths chapter 3 trigonometric functions. You will use the trigonometric identities in other chapters of mathematics and throughout the NCERT physics for both class 11 and 12 also. So it is very important to memorize and understand the basic identities and properties of trigonometric functions. CBSE NCERT solutions for class 11 maths chapter 3 trigonometric functions will help you for the same. Command on this chapter 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 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. There are four exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:3.1

Exercise:3.2

Exercise:3.3

Exercise:3.4

Miscellaneous Exercise

Let's understand this chapter with one example.

If we want to measure the height of a building such that we are standing at 50m away from building at point P and the angle of elevation made with the ground at P is 45 degree (as shown in the below figure). Then what will be the height of the building? Let the height of the building be QR and distance from the base of the building to point P is 50 meter

Using the trigonometric function

There are many other examples such as trigonometry is used in electric circuit analysis, predicting the heights of tides in the ocean, analyzing a musical tone and in seismology, etc.

## The main topics of this chapter are listed below:

3.1 Introduction

3.2 Angles

3.3 Trigonometric Functions

3.4 Trigonometric Functions of Sum and Difference of Two Angles

3.5 Trigonometric Equations

## NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.1

(i)
(ii)
(iii)
(iv)

It is solved using relation between degree and radian

(i)

(ii)
We know that

(iii)
We know that

(iv)
We know that

(1)

We know that

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  )

(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,

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

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

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

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

(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,

## Question:1 Find the values of other five trigonometric functions   , x lies in third quadrant.

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

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

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

Solution

lies in fourth quadrant. Therefore sin x is negative

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

We know that values of sin x repeat after an interval of

We know that value of cosec x repeats after an interval of

or

We know that tan x repeats after an interval of  or 180 degree

We know that sin x repeats after an interval of

We know that cot x repeats after an interval of

## CBSE NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.3

Question:1 Prove that

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:

= R.H.S.

Question:2 Prove that

The solutions for the given problem is done as follows.

R.H.S.

Question:3 Prove that

We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)

R.H.S.

Question:4 Prove that

Using the above values

R.H.S.

Question:5(i) Find the value of

We know that
(sin(x+y)=sinxcosy + cosxsiny)
Using this idendity

Question:5(ii) Find the value of

We know that,

By using this we can write

Question:6 Prove the following:

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:7 Prove the following

As we know that

and

So, by using these identities

R.H.S

Question:8 Prove the following

As we know that,
,      ,
and

By using these our equation simplify to

R.H.S.

Question:9 Prove the following

We know that

So, by using these our equation simplifies to

R.H.S.

Question:10 Prove the following

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

We know that

[ cos(A+B) - cos (A-B) = -2sinAsinB ]

By using this identity

R.H.S.

Question:12 Prove the following

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

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

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

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:16 Prove the following

As we know that

R.H.S.

Question:17 Prove the following

We know that

We use these identities

R.H.S.

Question:18 Prove the following

We know that

We use these identities

R.H.S.

Question:19 Prove the following

We know that

R.H.S.

Question:20 Prove the following

We know that

We use these  identities

R.H.S.

Question:21 Prove the following

We know that

We use these identities

=RHS

Question:22 prove the following

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

We know that

and we can write tan 4x = tan 2(2x)
So,       =

=

=

=           = R.H.S.

Question:24 Prove the following

We know that

We use this in our problem
cos 4x = cos 2(2x)
=
=
=  = R.H.S.

Question:25 Prove the following

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 solutions for class 11 maths chapter 3 trigonometric function-Exercise: 3.4

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

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

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  =

We know that

and also
So,
and

So, the principal solutions are

Now,

Therefore, the general solution is

where

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

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

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

We know that

So,

either
tan2x = 0             or                     tan2x = -1                                                  (      )
2x =

Where n  Z

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

## Question:1 Prove that

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

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

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

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

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

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

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

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

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

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)

## NCERT solutions for class 11 mathematics

 chapter-1 NCERT solutions for class 11 maths chapter 1 Sets chapter-2 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 maths chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13 NCERT solutions for class 11 maths chapter 13 Limits and Derivatives chapter-14 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE NCERT solutions for class 11 maths chapter 15 Statistics chapter-16 NCERT solutions for class 11 maths chapter 16 Probability

## NCERT solutions for class 11- Subject wise

 Solutions of NCERT for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry Solutions of NCERT for Class 11 physics

The basic identities used in NCERT solutions for class 11 maths chapter 3 Trigonometric Functions are listed below

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 solutions for 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.