# NCERT Solutions for Class 11 Maths Chapter 3

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions in English Medium Trikonmitiy Falan in Hindi Medium download in PDF file format to use it offline or use online without downloading for session 2020-2021.

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## NCERT Solutions for Class 11 Maths Chapter 3

 Class: 11 Maths (English and Hindi Medium) 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.

#### Class 11 Maths Chapter 3 Exercise 3.1 Solution in Videos

Class 11 Maths Chapter 3 Exercise 3.1 Solution
Class 11 Maths Exercise 3.1 Solution in Hindi

#### Class 11 Maths Chapter 3 Exercise 3.2 Solution in Videos

Class 11 Maths Chapter 3 Exercise 3.2 Solution
Class 11 Maths Exercise 3.2 Solution in Hindi

#### Class 11 Maths Chapter 3 Exercise 3.3 Solution in Videos

Class 11 Maths Chapter 3 Exercise 3.3 Solution
Class 11 Maths Exercise 3.3 Solution in Hindi

#### Class 11 Maths Chapter 3 Exercise 3.4 Solution in Videos

Class 11 Maths Chapter 3 Exercise 3.4 Solution
Class 11 Maths Exercise 3.4 Solution in Hindi

#### Class 11 Maths Chapter 3 Exercise 3.5 Solution in Videos

Class 11 Maths Chapter 3 Exercise 3.5 Solution
Class 11 Maths Exercise 3.5 Solution in Hindi

#### Class 11 Maths Chapter 3 Miscellaneous Exercise Solution in Videos

Class 11 Maths Chapter 3 Miscellaneous Exercise
Class 11 Maths Miscellaneous Exercise 3 in Hindi

#### 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°.
##### Class 11 Maths Chapter 3 Important Questions for Practice
• 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°.
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°
Therefore, the angle formed in 6 revolutions
= 6 × 2π
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
We know that π radians =180°
Therefore,
= 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                         