# 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 chapter 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 memorise and understand the basic identities and properties of trigonometric functions. 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. Here you will get NCERT solutions for class 11 also.

**Latest ****: NEET exam preparation bothering you? Why wait, start preparing with NEET Online Preparation Program. Know More**

**Latest ****: Why settle for a classroom full of children when you can have your own personalized classroom? Start preparing with our JEE Main Online Preparation Program. Know More**

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?

Prepare Online for JEE Main

Crack JEE with Online Preparation Program

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, analysing 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 **

** Question:1 ** Find the radian measures corresponding to the following degree measures:

** 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 solutions for class 11 maths chapter 3 trigonometric functions-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 solutions for class 11 maths chapter 3 trigonometric functions-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:20 ** Prove the following

** Answer: **

We know that

We use these identities

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 solutions for class 11 maths chapter 3 trigonometric function-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

**NCERT solutions for class 11 maths chapter 3 trigonometric functions-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)

**NCERT solutions for class 11 mathematics **

**NCERT solutions for class 11- Subject wise **

NCERT solutions for class 11 biology |

NCERT solutions for class 11 maths |

NCERT solutions for class 11 chemistry |

NCERT solutions 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.

** Happy Reading !!! **

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

**Question: **What are important topics of the chapter Trigonometric Functions ?

**Answer: **

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.

**Question: **How does the NCERT solutions are helpful ?

**Answer: **

NCERT solutions will help the students, if the stuck while solving the NCERT problems. Also, it will give them the new ways to solve the NCERT problems.

**Question: **Which are the most difficult chapters in the NCERT class 11 maths ?

**Answer: **

Permutation and combination, trigonometry are considered to be difficult chapters by the most of students but with the regular practice students can get command on them also.

**Question: **Which is the official website of NCERT ?

**Answer: **

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

**Question: **Does CBSE provides the solutions of NCERT class 11 ?

**Answer: **

No, CBSE doesn’t provided NCERT solutions for any class or subject.

**Question: **Where can I find the complete solutions of NCERT class 11 maths ?

**Answer: **

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.

## Latest Articles

##### IEO Result 2020-2021 - Check SOF IEO Level 1 & 2 Merit List Here

IEO Result 2020-2021 - Science Olympiad Foundation will releas...

##### Kendriya Vidyalaya Bangalore Admission 2021 - Apply Here

Kendriya Vidyalaya Bangalore Admission 2021 - Know the admissi...

##### iOM Result 2020-21 - Check Silverzone iOM Level 1 & 2 Result ...

iOM Result 2020-21 - Silverzone International Olympiad of Mat...

##### IOEL Result 2020-21 - Check SilverZone IOEL Result Here

IOEL Result 2020-21 - SilverZone Foundation will release the r...

##### Kendriya Vidyalaya Pune Admission 2021 - Check Eligibility, Re...

Kendriya Vidyalaya Pune Admission 2021 - Check the admission p...

##### JAC 9th Syllabus 2020-2021 - Download Jharkhand Board Class 9t...

JAC Cass 9th Syllabus 2020-2021 - Students can download Jharkh...

##### JAC 11th Syllabus 2020-2021 - Check Jharkhand Board Class 11 S...

JAC 11th Syllabus 2020-2021 - The syllabus of Class 11 JAC for...

##### Rashtriya Military School Result 2021 Date - Check RMS Result ...

Rashtriya Military School Result 2021 - Rashtriya Military Sch...

##### IOS Result 2020-21 for Level 1 - Check Silverzone iOS Results ...

IOS Result 2020-21 - Silverzone Foundation will release Scienc...

##### KVPY Admit Card 2020-2021 - Download KVPY Hall Ticket Here

KVPY Admit Card 2020-2021 - IISc Bangalore will release the K...