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Have you ever wondered how engineers design huge buildings, how sailors and pilots navigate their journey, or how shadows change their length throughout the day? You can find all these answers in trigonometry, a fascinating branch of mathematics. In class 11 Maths NCERT chapter 3, trigonometric functions, contains the advanced concepts of trigonometry like radian measure, relation between degree and radian, sign of trigonometric functions, graphs of trigonometric functions, domain and range of trigonometric functions, and trigonometric functions of the sum and difference of two angles. Understanding these concepts will make students more efficient in solving problems involving height, distance, and angles. NCERT solutions of class 11 of trigonometric functions will also build a strong foundation for more advanced trigonometric concepts, which have many practical, real-life applications like construction, navigation, architecture, etc.
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This article on NCERT solutions for class 11 Maths Chapter 3 Trigonometric Functions offers clear and step-by-step solutions for the exercise problems in the NCERT Class 11 Maths Book. Students requiring trigonometric functions class 11 solutions will find this article useful. It covers all the important Class 11 Maths Chapter 3 question answers and concepts of trigonometric functions. According to the latest CBSE syllabus, the experienced Subject Matter Experts of Careers360 have curated these trigonometric functions class 11 NCERT solutions, ensuring that students can grasp the basic concepts effectively. For syllabus, notes, and PDF, refer to this link: NCERT
NCERT Trigonometric Functions Class 11 Solutions: Exercise: 3.1 Page Number: 48-49 Total Questions: 7 |
Question 1: Find the radian measures corresponding to the following degree measures:
(i)
(ii)
(iii)
(iv)
Answer:
It is solved using the relation between degree and radian
(i)
We know that
So,
(ii)
We know that
Now, we know that
So,
(iii)
We know that
So,
(iv)
We know that
So,
Question 2: Find the degree measures corresponding to the following radian measures. (Use
Answer:
(i)
We know that
So,
(we use
Here 1' represents 1 minute and 60" represents 60 seconds
Now,
(ii)
We know that
So,
(we use
(iii)
We know that
So,
(iv)
We know that
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
(
In one revolution, the wheel will cover
So, in 6 revolutions it will cover =
Answer:
We know that
here
and
Now,
We know that
So, 1 radian =
So,
Thus, the angle subtended at the centre of a circle
Question 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.
Answer:
Given: radius (
length of chord = 20 cm
We know that
Now,
AB is the chord of length 20 cm and OA and OB are radii of circle i.e. 20 cm each
The angle subtended by OA and OB at centre =
So, each angle is
Now, we have
So,
Answer:
Given:
We need to find the ratio of their radii
We know that arc length
So,
Now,
So,
Answer:
(i) We know that
Now,
So,
(ii) We know that
Now,
So,
(iii) We know that
Now,
So,
NCERT Trigonometric Functions Class 11 Solutions: Exercise: 3.2 Page Number: 57 Total Questions: 10 |
Question 1: Find the values of other five trigonometric functions as
Answer:
Question 2: Find the values of other five trigonometric functions as
Answer:
Question 3: Find the values of other five trigonometric functions as
Answer:
Question 4: Find the values of other five trigonometric functions as
Answer:
Question 5: Find the values of the other five trigonometric functions as
Answer:
Therefore the value of
Question 6: Find the values of the trigonometric functions
Answer:
We know that values of
Question 7: Find the values of the trigonometric functions
Answer:
We know that value of
Question 8: Find the values of the trigonometric functions
Answer:
We know that
Question 9: Find the values of the trigonometric functions
Answer:
We know that
Question 10: Find the values of the trigonometric functions
Answer:
We know that
NCERT Trigonometric Functions Class 11 Solutions: Exercise: 3.3 Page Number: 67-68 Total Questions: 25 |
Question 1: Prove that
Answer:
We know the values of sin 30°, cos 60°, and tan 45°.
L.H.S.
Question 2: Prove that
Answer:
L.H.S. =
= R.H.S.
Question 3: Prove that
Answer:
We know the values of cot 30°, tan 30°, and cosec 30°.
L.H.S.
Question 4: Prove that
Answer:
Using the above values
L.H.S.
Question 5(i): Find the value of
Answer:
We know that
Using this idendity
Question 5 (ii): Find the value of
Answer:
We know that,
By using this we can write,
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
Question 7: Prove the following
Answer:
As we know that
So, by using these identities
L.H.S.
Question 8: Prove the following
Answer:
As we know,
By using these our equation simplify to
Question 9: Prove the following
Answer:
We know that
So, by using these our equation simplifies to
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)
Now,
Question 11: Prove the following
Answer:
We know that
[ cos(A+B) - cos (A-B) = -2sinAsinB ]
By using this identity
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
So,
Question 13: Prove the following
Answer:
As we know that
Now,
By using these identities
cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x (
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x
So our equation becomes
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) (
So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) (
=
=
=
Question 15: Prove the following
Answer:
We know that
By using this , we get
sin5x + sin3x = 2sin4xcosx
Now multiply and divide by sin x
Now we know that
By using this our equation becomes
Question 16: Prove the following
Answer:
As we know that
Now,
Question 17: Prove the following
Answer:
We know that
We use these identities
Now,
Question 18: Prove the following
Answer:
We know that
We use these identities
Question 19: Prove the following
Answer:
We know that
We use these equations,
Now,
Question 20: Prove the following
Answer:
We know that
We use these identities
Now,
Question 21: Prove the following
Answer:
We know that
We use these identities
Question 22: prove the following
Answer:
L.H.S.
=
Now we can write
⇒
So,
=
=
Question 23: Prove that
Answer:
We know that
and we can write tan 4x = tan 2(2x)
So,
=
=
=
Question 24: Prove the following
Answer:
We know that
We use this in our problem
=
=
=
Question 25: Prove the following
Answer:
We know that
We use this in our problem
we can write
=
=
=
=
NCERT Trigonometric Functions Class 11 Solutions: Miscellaneous Exercise Page Number: 71-72 Total Questions: 10 |
Question 1: Prove that
Answer:
We know that
cos A+ cos B =
we use this in our problem
=
=
=
again use the above identity
=
=
we know that
So,
Question 2: Prove that
Answer:
We know that
We use this in our problem
=
= (4sinx - 4
Now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4
= 4
= 0 = R.H.S.
Question 3: Prove that
Answer:
We know that
We use these two in our problem
= 1 + 2 cos x cos y + 1 - 2 sin x sin y
= 2 + 2(cos x cos y - sin x sin y)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
=
=
= R.H.S.
Question 4: Prove that
Answer:
We know that
We use these two in our problem
= 1 - 2cos x cos y + 1 - 2sin x sin y
= 2 - 2(cos x cos y + sin x sin y)
= 2 - 2cos(x - y)
= 2(1 - cos(x - y) )
Now we can write
So,
=
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 need a combination of sin7x and sin x , sin3x and sin5x to get sin4x.
take 2sin4x common
= 2sin4x(cos3x + cosx)
We know that
We use this
=
Now 2sin4x(cos3x + cosx) = 2sin4x(
=
Question 6: Prove that
Answer:
We know that
We use these two identities in our problem
sin7x + sin5x =
sin 9x + sin 3x =
cos 7x + cos5x =
cos 9x + cos3x =
=
Question:7: Prove that
Answer:
We know that
We use these identities
sin2x +
take 2 sinx common
Question 8: Find
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
we know that
⇒
⇒
x lies in II quadrant So value of sin x is +ve
Question 9: Find
Answer:
We know that
cos x =
⇒
⇒
we know that
cos x =
⇒
⇒
Because
Question 10: Find
Answer:
We know that
cos x =
We know that
cosx =
Similarly,
cos x =
Also, read,
Question:
The solution of
Solution:
Hence, the correct answer is
The topics discussed in the NCERT Solutions for class 11, chapter 3, Trigonometric Functions are:
Radian Measure =
Degree Measure =
sin θ = P / H
cos θ = B / H
tan θ = P / B
cosec θ = H / P
sec θ = H / B
cot θ = B / P
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
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
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
sin² θ + cos² θ = 1
cosec² θ – cot² θ = 1
sec² θ – tan² θ = 1
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)]
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]
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
tan (x+y) = (tan x + tan y) / (1 − tan x tan y)
tan (x−y) = (tan x − tan y) / (1 + tan x tan y)
tan 2θ = (2 tan θ) / (1 - tan²θ)
sin 3θ = 3sin θ – 4sin³θ
cos 3θ = 4cos³θ – 3cos θ
tan 3θ = [3tan θ – tan³θ] / [1 − 3tan²θ]
Here are some approaches that students can use to approach the questions related to trigonometric functions.
Here is a comparison list of the concepts in Trigonometric Functions that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:
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Trigonometric Equation using Minimum and Maximum value of Function |
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Given below is the chapter-wise list of the NCERT Class 11 Maths solutions with their respective links:
Also, read,
Here are the subject-wise links for the NCERT solutions of class 11:
Here are some useful links for NCERT books and the NCERT syllabus for class 11:
To solve trigonometric equations in Class 11 Maths Chapter 3, express the given terms in basic trigonometric functions like sin, cos, and tan, then try to apply proper trigonometric identities and use simplification.
The key formulas of Trigonometric Functions in NCERT Class 11 are:
To find the general solution of a trigonometric equation, use the following standard equations:
In class 11 chapter 3 trigonometric functions, a domain basically denotes the input values and range denotes the output values.
For example: The domain of
To derive the sine and cosine function graphs, plot
The sine graph starts at (0, 0) and goes up to 1 at 90°, drops to –1 at 270°, and comes to 0 again at 360° then repeats.
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