NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions : Trigonometry has various realtime applications. It is used to solve height and distance problems. This cahpter gives an introduction to basic properties and identities of trigonom etric 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.
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 functionsExercise: 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,
Solutions for class 11 maths chapter 3 trigonometric functionsExercise: 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
CBSE NCERT solutions for class 11 maths chapter 3 trigonometric functionsExercise: 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(AB) (i)
2sinAsinB = cos(A+B)  cos(AB) (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(AB)
2cosAcosB = cos(A+B) + cos(AB)
R.H.S.
Question:11 Prove the following
Answer:
We know that
[ cos(A+B)  cos (AB) = 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(AB)
2cosAsinB = sin(A+B)  sin(AB)
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 functionExercise: 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(2cosx1) = 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
Solutions of NCERT for class 11 maths chapter 3 trigonometric functionsMiscellaneous 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
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 !!!