NCERT Solutions for Class 11 Maths Chapter 3

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions त्रिकोणमितीय फलन in Hindi & English Medium download in PDF form to use it offline or use online without downloading for session 2020-21.

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NCERT Solutions for Class 11 Maths Chapter 3

Class:11
Subject:Maths – गणित
Chapter 3:Trigonometric Functions

11th Maths Chapter 3 Solutions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are given below to download in PDF or use online in Hindi and English Medium. Contents are updated for academic session 2020-21 for UP Board, MP Board, CBSE and all other boards who are using NCERT Books 2020-21 as their course books.



Questions for Practice

1. Write the value of 2sin 75° sin 15°?
2. What is the maximum value of 3 – 7 cos 5x?
3. Express sin 12A + sin 4A as the product of sines and cosines.
4. Express 2 cos 4x sin 2x as an algebraic sum of sines and cosines
5. Write the maximum value of cos (cos x).
6. Write the minimum value of cos (cos x).
7. Write the radian measure of 22° 30’
8. Find the length of an arc of a circle of radius 5cm subtending a central angle measuring 15°.




Important Questions

1. Find the maximum and minimum value of 7 cos x + 24 sin x
2. Evaluate sin(π + x) sin(π – x) cosec² x
3. Find the angle in radians between the hands of a clock at 7 : 20 pm.
4. A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 metres when it traces 72° at the centre, find the length of the rope.
5. Draw sin x, sin 2x and sin 3x on same graph and with same scale.



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Important Questions on 11th Maths Chapter 3

Find the radian measures corresponding to 25°.
We know that 180°=π radians
Therefore, 25° = π/180×25 radians
=5π/36 radians
Hence, 25°=5π/36 radians
मान ज्ञात कीजिए: sin⁡〖765°〗
sin⁡〖765°〗
=sin⁡(2×360° + 45°)
〖=sin 45°〗
[∵ पहले चतुर्थांश में sin धनात्मक होता है]
=1/√2
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Number of revolutions in one minute (60 seconds) =360
Therefore, number of revolutions in 1 seconds
= 360/60
= 6
We know that the angle formed in one revolutions
= 360°
= 2π radians
Therefore, the angle formed in 6 revolutions
= 6 × 2π
= 12π radians
Hence, it will turn 12π radians in one second.
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).
Here, radius r=100 cm, length of arc l=22 cm
Hence, using the relation θ=l/r we have
θ=22/100 radians =11/50 radians
We know that π radians =180°
Therefore,
11/50 radians
= 180/π×11/50 degree
= (180×7)/22×11/50 degree
= 63/5 degree
= 12 3/5 degree
= 12° + 3/5 × 60 minutes
[∵1°=60′]
= 12° + 36 minutes
= 12° + 36’= 12°36′
Hence, the angle formed by are at the centre is 12°36′.
सिद्ध कीजिए: 2 cos⁡〖π/13〗 cos⁡〖9π/13〗+cos⁡〖3π/13〗+cos⁡〖5π/13〗= 0
बायाँ पक्ष =2 cos⁡〖π/13〗 cos⁡〖9π/13〗+cos⁡〖3π/13〗+cos⁡〖5π/13〗
= cos⁡(π/13+9π/13)+cos⁡(π/13-9π/13)+cos⁡〖3π/13〗+cos⁡〖5π/13〗
[∵2 cos⁡A cos⁡B=cos⁡(A+B)+cos⁡(A-B) ]
= cos⁡〖10π/13〗+cos⁡〖8π/13〗+cos⁡〖3π/13〗+cos⁡〖5π/13〗
= cos⁡(π-3π/13)+cos⁡(π-5π/13)+cos⁡〖3π/13〗+cos⁡〖5π/13〗
= -cos⁡〖3π/13〗-cos⁡〖5π/13〗+cos⁡〖3π/13〗+cos⁡〖5π/13〗
= 0 = दायाँ पक्ष
सिद्ध कीजिए: (sin⁡3x + sin⁡x ) sin⁡x + (cos⁡3x – cos⁡x ) cos⁡x = 0
बायाँ पक्ष
= (sin⁡3x+sin⁡x ) sin⁡x + (cos⁡3x-cos⁡x) cos⁡x
=(2 sin⁡〖(3x+x)/2〗 cos⁡〖(3x-x)/2〗) sin⁡x+(-2 sin⁡〖(3x+x)/2〗 sin⁡〖(3x-x)/2〗 ) cos⁡x
[∵sin⁡A+sin⁡B=2 sin⁡〖(A+B)/2〗 cos⁡〖(A-B)/2〗 तथा cos⁡A-cos⁡B=-2 sin⁡〖(A+B)/2〗 sin⁡〖(A-B)/2〗 ]
= (2 sin⁡2x cos⁡x ) sin⁡x+(-2 sin⁡2x sin⁡x ) cos⁡x
= 2 sin⁡2x sin⁡x cos⁡x-2 sin⁡2x sin⁡x cos⁡x
= 0
= दायाँ पक्ष
Prove that: sin^2⁡〖π/6〗+cos^2⁡〖π/3〗-tan^2⁡〖π/4〗= -1/2
LHS
= sin^2⁡〖π/6〗+cos^2⁡〖π/3〗-tan^2⁡〖π/4〗
= (1/2)^2 + (1/2)^2 – (1)^2
= 1/4 + 1/4 – 1
= 1/2 – 1
= – 1/2
= RHS
Prove that: sin⁡(n+1)x sin⁡(n + 2)x + cos⁡(n+1)x cos⁡(n+2)x = cos⁡x
LHS
= sin⁡(n+1)x sin⁡(n+2)x + cos⁡(n+1)x cos⁡(n+2)x
=cos⁡[(n+2)x-(n+1)x]
[∵cos⁡A cos⁡B + sin⁡A sin⁡B = cos⁡(A-B) ]
= cos⁡[nx+2x-nx-x]
= cos⁡x
= RHS
सिद्ध कीजिए: sin⁡x + sin⁡3x + sin⁡5x + sin⁡7x =4 cos⁡x cos⁡2x sin⁡4x
बायाँ पक्ष
= sin⁡x + sin⁡3x + sin⁡5x + sin⁡7x
= (sin⁡7x + sin⁡x) + (sin⁡5x + sin⁡3x)
= 2 sin⁡〖(7x + x)/2〗cos⁡〖(7x – x)/2〗+ 2sin⁡〖(5x + 3x)/2〗cos⁡〖(5x – 3x)/2〗
[∵ sin⁡A +sin⁡B = 2sin⁡((A+B)/2) cos⁡((A-B)/2) ]
= 2 sin⁡4x cos⁡3x+2 sin⁡4x cos⁡x
= 2 sin⁡4x (cos⁡3x+cos⁡x )
= 2 sin⁡4x (2 cos⁡〖(3x+x)/2〗 cos⁡〖(3x-x)/2〗 )
[∵ cos⁡A + cos⁡B =2cos⁡((A + B)/2) cos⁡((A – B)/2) ]
= 2 sin⁡4x (2 cos⁡2x cos⁡x )
= 4 cos⁡x cos⁡2x sin⁡4x
= दायाँ पक्ष
Prove that: cos⁡4x = 1 – 8 sin^2⁡x cos^2⁡x
LHS
= cos⁡4x
= cos⁡2(2x)
= 1 – 2sin^2(⁡2x)
= 1-2(sin⁡2x )^2
[∵cos⁡2A=1-2 sin^2⁡A ]
= 1 – 2(2sin⁡x cos⁡x )^2
[∵sin⁡2A = 2 sin⁡A cos⁡A ]
= 1 – 2(4sin^2⁡x cos^2⁡x )
= 1 – 8sin^2⁡x cos^2⁡x
= RHS
Prove that: cos⁡6x = 32 cos^6⁡x – 48 cos^4⁡x + 18 cos^2⁡x – 1
LHS
= cos⁡6x
= cos⁡2(3x)
= 2cos^2(⁡2x) – 1
[∵ cos⁡2A =2 cos^2(A) – 1]
= 2(4 cos^3⁡x-3 cos⁡x )^2 – 1
[∵ cos⁡3A = 4 cos^3 A – 3 cos A]
= 2(16 cos^6⁡x+9 cos^2⁡x-24 cos^4⁡x )-1
= 32 cos^6⁡x – 48 cos^4⁡x + 18 cos^2⁡x – 1
= RHS

11 Maths Chapter 3 Exercise 3.1 Question 1, 2, 3
11 Maths Chapter 3 Exercise 3.1 Question 4, 5
11 Maths Chapter 3 Exercise 3.1 Question 6, 7 in English medium
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2 in English medium
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2 updated for 2018-19.
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2 for up board and cbse.
11 Maths Chapter 3 Exercise 3.3 solutions in English
11 Maths Chapter 3 Exercise 3.3 solutions for 2018-19
11 Maths Chapter 3 Exercise 3.3 solutions updated for up board
11 Maths Chapter 3 Exercise 3.3 solutions questions 1, 2, 3, 4, 5, 6, 7.
11 Maths Chapter 3 Exercise 3.3 solutions question 8, 9, 10, 11, 12, 13, 14.
11 Maths Chapter 3 Exercise 3.3 solutions Questions 15, 16, 17, 18, 19, 20.
11 Maths Chapter 3 Exercise 3.3 solutions Questions 21, 22, 23, 24, 25.
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.4 in English
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.4 for 2018-19.
NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.4 sols for up and gujrat board.
Class 11 Maths Chapter 3 Exercise 3.5 Supplementary Exercise in English
Class 11 Maths Chapter 3 Exercise 3.5 Supplementary Exercise all questions
Class 11 Maths Chapter 3 Exercise 3.5 Supplementary Exercise question 1, 2, 3, 4, 5
Class 11 Maths Chapter 3 Exercise 3.5 Supplementary question 6, 7, 8, 9, 10, 11, 12, 13
Class 11 Maths Chapter 3 Exercise 3.5 Supplementary Exercise for 2018-19.
NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise for 2018-19.
NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise all questions in updated form.
NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise for all board english medium.
NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise in PDF.