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NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q1:  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: L.H.S. =  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q2: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: L.H.S. =  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q3: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: L.H.S. = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q4: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: L.H.S = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q5: Find the value of:
(i) sin 75°  
(ii) tan 15°
Ans:  (i) sin 75° = sin (45°+ 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x  + y) = sin x cos y  + cos x sin y]

  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
(ii) tan 15° = tan (45° – 30°)
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q6: Prove that: NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: 
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q7: Prove that:  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
∴ L.H.S. = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q8: Prove that NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: 
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q9: NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: L.H.S. =  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q10: Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Ans: L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q11: Prove that NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions .

∴L.H.S. =  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

             NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q12: Prove that sin2 6x – sin2 4x = sin 2x sin 10x
Ans: It is known that 

 NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
∴ L.H.S. = sin26x – sin24x
= (sin 6x + sin 4x) (sin 6x – sin 4x)  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S.

Q13: Prove that cos2 2x – cos2 6x = sin 4sin 8x
Ans: It is known that 

 NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
∴ L.H.S. = cos2 2x – cos2 6x
= (cos 2x + cos 6x) (cos 2– 6x)

NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
= [2 cos 4x cos 2x] [–2 sin 4(–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.

Q14: Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Ans: L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] +2 sin 4x

NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
= 2 sin 4x cos (– 2x)+ 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos2 x – 1+ 1)
= 2 sin 4x (2 cos2 x)
= 4cos2 x sin 4x
= R.H.S.

Q15: Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Ans: L.H.S = cot 4x (sin 5x  sin 3x)
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
= 2 cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
= 2 cos 4x. cos x
L.H.S. = R.H.S.

Q16:  Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
∴ L.H.S = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q17: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that

  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

∴L.H.S. = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

              NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q18: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that

NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
∴ L.H.S. = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

               NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q19: Prove that   NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that

NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

∴L.H.S. = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

              NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q20: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that

NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

∴L.H.S. =  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

              NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q21: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: L.H.S. = NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q22: Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Ans: L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
= cot x cot 2x – cot 3x (cot 2x + cot x)
= cot x cot 2x – cot (2 + x) (cot 2x + cot x)
NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
= cot x cot 2– (cot 2cot x – 1)
= 1 = R.H.S.

Q23: Prove that  NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions
Ans: It is known that NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions .
∴ L.H.S. = tan 4x = tan 2(2x)

              NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

Q24: Prove that cos 4x = 1 – 8sincosx
Ans: L.H.S. = cos 4x
= cos 2(2x)
= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]
= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]
= 1 – 8 sin2x cos2x
= R.H.S.

Q25: Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 – 1
Ans: L.H.S. = cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]
= 4 [(2 cos2 – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 – 1]
= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3
= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3
= 32 cos6– 48 cos4x + 18 cos2x – 1
= R.H.S.

The document NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on NCERT Solutions Class 11 Maths Chapter 3 - Trigonometric Functions

1. What are the basic trigonometric functions?
Ans. The basic trigonometric functions are sine, cosine, and tangent, which are defined as ratios of the sides of a right triangle.
2. How are trigonometric functions used in real-life applications?
Ans. Trigonometric functions are used in various fields such as engineering, physics, and astronomy to analyze and solve problems involving angles and distances.
3. What is the unit circle and how is it related to trigonometric functions?
Ans. The unit circle is a circle with a radius of 1 centered at the origin. Trigonometric functions can be defined based on the coordinates of points on the unit circle.
4. How do you find the values of trigonometric functions for angles beyond 90 degrees?
Ans. Trigonometric functions for angles beyond 90 degrees can be found using reference angles and the periodic nature of trigonometric functions.
5. Can trigonometric functions be used to solve for unknown sides and angles in a triangle?
Ans. Yes, trigonometric functions such as sine, cosine, and tangent can be used to solve for unknown sides and angles in a right triangle through trigonometric ratios.
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