# 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.

### 11th Maths Chapter 3 Solutions in English Medium

- Class 10 Maths Exercise 3.1 Solutions in English
- Class 10 Maths Exercise 3.2 Solutions in English
- Class 10 Maths Exercise 3.3 Solutions in English
- Class 10 Maths Exercise 3.4 Solutions in English
- Class 10 Maths Exercise 3.5 (Supplementary) Solutions
- Class 10 Maths Miscellaneous Exercise 3 Solution in English

### 11th Maths Chapter 3 Solutions in Hindi Medium

- Class 10 Maths Exercise 3.1 Solutions in Hindi
- Class 10 Maths Exercise 3.2 Solutions in Hindi
- Class 10 Maths Exercise 3.3 Solutions in Hindi
- Class 10 Maths Exercise 3.4 Solutions in Hindi
- Class 10 Maths Exercise 3.5 Solutions in Hindi
- Class 10 Maths Miscellaneous Exercise 3 Solutions in Hindi
- Class 11 Maths Solutions Main Page

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

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

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

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

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

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

#### Questions for Practice

- Write the value of 2sin 75° sin 15°?
- What is the maximum value of 3 – 7 cos 5x?
- Express sin 12A + sin 4A as the product of sines and cosines.
- Express 2 cos 4x sin 2x as an algebraic sum of sines and cosines
- Write the maximum value of cos (cos x).
- Write the minimum value of cos (cos x).
- Write the radian measure of 22° 30’
- 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.

###### Feedback & Suggestions

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### Important Questions on 11th Maths Chapter 3

Therefore, 25° = π/180×25 radians

=5π/36 radians

Hence, 25°=5π/36 radians

=sin(2×360° + 45°)

〖=sin 45°〗

[∵ पहले चतुर्थांश में sin धनात्मक होता है]

=1/√2

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.

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′.

= cos(π/13+9π/13)+cos(π/13-9π/13)+cos〖3π/13〗+cos〖5π/13〗

[∵2 cosA cosB=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

= (sin3x+sinx ) sinx + (cos3x-cosx) cosx

=(2 sin〖(3x+x)/2〗 cos〖(3x-x)/2〗) sinx+(-2 sin〖(3x+x)/2〗 sin〖(3x-x)/2〗 ) cosx

[∵sinA+sinB=2 sin〖(A+B)/2〗 cos〖(A-B)/2〗 तथा cosA-cosB=-2 sin〖(A+B)/2〗 sin〖(A-B)/2〗 ]

= (2 sin2x cosx ) sinx+(-2 sin2x sinx ) cosx

= 2 sin2x sinx cosx-2 sin2x sinx cosx

= 0

= दायाँ पक्ष

= 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

= sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x

=cos[(n+2)x-(n+1)x]

[∵cosA cosB + sinA sinB = cos(A-B) ]

= cos[nx+2x-nx-x]

= cosx

= RHS

= sinx + sin3x + sin5x + sin7x

= (sin7x + sinx) + (sin5x + sin3x)

= 2 sin〖(7x + x)/2〗cos〖(7x – x)/2〗+ 2sin〖(5x + 3x)/2〗cos〖(5x – 3x)/2〗

[∵ sinA +sinB = 2sin((A+B)/2) cos((A-B)/2) ]

= 2 sin4x cos3x+2 sin4x cosx

= 2 sin4x (cos3x+cosx )

= 2 sin4x (2 cos〖(3x+x)/2〗 cos〖(3x-x)/2〗 )

[∵ cosA + cosB =2cos((A + B)/2) cos((A – B)/2) ]

= 2 sin4x (2 cos2x cosx )

= 4 cosx cos2x sin4x

= दायाँ पक्ष

= cos4x

= cos2(2x)

= 1 – 2sin^2(2x)

= 1-2(sin2x )^2

[∵cos2A=1-2 sin^2A ]

= 1 – 2(2sinx cosx )^2

[∵sin2A = 2 sinA cosA ]

= 1 – 2(4sin^2x cos^2x )

= 1 – 8sin^2x cos^2x

= RHS

= cos6x

= cos2(3x)

= 2cos^2(2x) – 1

[∵ cos2A =2 cos^2(A) – 1]

= 2(4 cos^3x-3 cosx )^2 – 1

[∵ cos3A = 4 cos^3 A – 3 cos A]

= 2(16 cos^6x+9 cos^2x-24 cos^4x )-1

= 32 cos^6x – 48 cos^4x + 18 cos^2x – 1

= RHS