NCERT Solutions for Class 11 Maths Chapter 3

Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions त्रिकोणमितीय फलन in Hindi & English Medium download in PDF form to use it offline. Download Offline Apps based on latest NCERT Solutions following updated CBSE Syllabus for 2019-20.


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

NCERT Solutions for Class 11 Maths Chapter 3

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11 Maths Chapter 3 Trigonometric Functions Solutions

Hindi Medium and English Medium




Important Extra Questions on Trigonometric Functions
  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°.



Try These
  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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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

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