Class 11 Mathematics Notes Chapter 3 (Chapter 3) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 3, Trigonometric Functions, from your NCERT Exemplar. This is a crucial chapter, not just for your Class 11 exams, but it forms the backbone for much of calculus and is frequently tested in various government entrance examinations. We'll break down the key concepts methodically.

Chapter 3: Trigonometric Functions - Detailed Notes

1. Angles:

• Concept: An angle is a measure of rotation of a given ray about its initial point. The original ray is the initial side, and the final position is the terminal side.
• Units of Measurement:
  • Degree Measure (Sexagesimal System): 1 right angle = 90 degrees (90°). 1° = 60 minutes (60'). 1' = 60 seconds (60").
  • Radian Measure (Circular System): An angle subtended at the centre by an arc of length equal to the radius is 1 radian (1ᶜ). It's a constant angle.
• Relation between Degree and Radian:
  • A complete circle subtends 2π radians or 360°.
  • π radians = 180°
  • Conversion:
    • Radian measure = (π / 180) × Degree measure
    • Degree measure = (180 / π) × Radian measure
  • Approximate value: 1 radian ≈ 57° 16' and 1° ≈ 0.01746 radians.
• Relation between Arc Length (l), Radius (r), and Angle (θ in radians):
  • l = rθ (θ must be in radians)

2. Trigonometric Functions:

• Unit Circle Definition: Consider a unit circle (radius = 1, centre at origin). Let P(a, b) be any point on the circle such that the angle made by OP with the positive x-axis is θ.

  • cos θ = a (x-coordinate)
  • sin θ = b (y-coordinate)
  • Since P(a, b) lies on x² + y² = 1, we get the fundamental identity: cos² θ + sin² θ = 1

• Other Trigonometric Functions (in terms of sin and cos):

  • tan θ = sin θ / cos θ (provided cos θ ≠ 0)
  • cot θ = cos θ / sin θ (provided sin θ ≠ 0)
  • sec θ = 1 / cos θ (provided cos θ ≠ 0)
  • cosec θ = 1 / sin θ (provided sin θ ≠ 0)

• Signs in Quadrants (ASTC Rule):

  • Quadrant I (0 < θ < π/2): All positive (All)
  • Quadrant II (π/2 < θ < π): Sine (and cosec) positive (Sin)
  • Quadrant III (π < θ < 3π/2): Tangent (and cot) positive (Tan)
  • Quadrant IV (3π/2 < θ < 2π): Cosine (and sec) positive (Cos)

• Domain and Range:

  Function	Domain	Range
  sin θ	R (all real numbers)	[-1, 1]
  cos θ	R	[-1, 1]
  tan θ	R - { (2n+1)π/2	n ∈ Z }
  cot θ	R - { nπ	n ∈ Z }
  sec θ	R - { (2n+1)π/2	n ∈ Z }
  cosec θ	R - { nπ	n ∈ Z }

• Periodicity:

  • sin(θ + 2nπ) = sin θ
  • cos(θ + 2nπ) = cos θ
  • cosec(θ + 2nπ) = cosec θ
  • sec(θ + 2nπ) = sec θ (Period is 2π)
  • tan(θ + nπ) = tan θ
  • cot(θ + nπ) = cot θ (Period is π)

3. Trigonometric Identities:

• Fundamental Identities:
  • sin² θ + cos² θ = 1
  • 1 + tan² θ = sec² θ
  • 1 + cot² θ = cosec² θ
• Negative Angle Identities:
  • sin(-θ) = -sin θ
  • cos(-θ) = cos θ
  • tan(-θ) = -tan θ
  • cot(-θ) = -cot θ
  • sec(-θ) = sec θ
  • cosec(-θ) = -cosec θ
• Allied Angle Identities (Examples): (Use ASTC rule and function changes for π/2, 3π/2)
  • sin(π/2 - θ) = cos θ, cos(π/2 - θ) = sin θ, tan(π/2 - θ) = cot θ
  • sin(π - θ) = sin θ, cos(π - θ) = -cos θ, tan(π - θ) = -tan θ
  • sin(π + θ) = -sin θ, cos(π + θ) = -cos θ, tan(π + θ) = tan θ
  • sin(2π - θ) = -sin θ, cos(2π - θ) = cos θ, tan(2π - θ) = -tan θ
• Sum and Difference Formulas:
  • cos(A + B) = cos A cos B - sin A sin B
  • cos(A - B) = cos A cos B + sin A sin B
  • sin(A + B) = sin A cos B + cos A sin B
  • sin(A - B) = sin A cos B - cos A sin B
  • tan(A + B) = (tan A + tan B) / (1 - tan A tan B)
  • tan(A - B) = (tan A - tan B) / (1 + tan A tan B)
  • cot(A + B) = (cot A cot B - 1) / (cot B + cot A)
  • cot(A - B) = (cot A cot B + 1) / (cot B - cot A)
• Multiple Angle Formulas (Double Angle):
  • sin 2A = 2 sin A cos A = (2 tan A) / (1 + tan² A)
  • cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A = (1 - tan² A) / (1 + tan² A)
  • tan 2A = (2 tan A) / (1 - tan² A)
• Multiple Angle Formulas (Triple Angle):
  • sin 3A = 3 sin A - 4 sin³ A
  • cos 3A = 4 cos³ A - 3 cos A
  • tan 3A = (3 tan A - tan³ A) / (1 - 3 tan² A)
• Product-to-Sum Formulas:
  • 2 sin A cos B = sin(A + B) + sin(A - B)
  • 2 cos A sin B = sin(A + B) - sin(A - B)
  • 2 cos A cos B = cos(A + B) + cos(A - B)
  • 2 sin A sin B = cos(A - B) - cos(A + B) (Note the order)
• Sum-to-Product Formulas (C-D Formulas):
  • sin C + sin D = 2 sin((C+D)/2) cos((C-D)/2)
  • sin C - sin D = 2 cos((C+D)/2) sin((C-D)/2)
  • cos C + cos D = 2 cos((C+D)/2) cos((C-D)/2)
  • cos C - cos D = -2 sin((C+D)/2) sin((C-D)/2) = 2 sin((C+D)/2) sin((D-C)/2) (Note the negative sign or angle swap)
• Half-Angle Formulas (less common but useful):
  • sin(A/2) = ±√[(1 - cos A) / 2]
  • cos(A/2) = ±√[(1 + cos A) / 2]
  • tan(A/2) = ±√[(1 - cos A) / (1 + cos A)] = sin A / (1 + cos A) = (1 - cos A) / sin A
  • (Sign depends on the quadrant of A/2)

4. Trigonometric Equations:

• Principal Solutions: Solutions lying in the interval [0, 2π).
• General Solutions: Expression involving integer 'n' (n ∈ Z) giving all possible solutions.
  • If sin θ = sin α ⇒ θ = nπ + (-1)ⁿ α, n ∈ Z
  • If cos θ = cos α ⇒ θ = 2nπ ± α, n ∈ Z
  • If tan θ = tan α ⇒ θ = nπ + α, n ∈ Z
  • If sin² θ = sin² α ⇒ θ = nπ ± α, n ∈ Z
  • If cos² θ = cos² α ⇒ θ = nπ ± α, n ∈ Z
  • If tan² θ = tan² α ⇒ θ = nπ ± α, n ∈ Z
• Solving Strategy:
  1. Simplify the equation using identities.
  2. Reduce it to one of the basic forms (sin θ = sin α, etc.).
  3. Find the principal value α.
  4. Write the general solution using the appropriate formula.

5. Maximum and Minimum Values:

• The maximum value of a cos θ + b sin θ is √(a² + b²)
• The minimum value of a cos θ + b sin θ is -√(a² + b²)
• Remember the ranges: -1 ≤ sin θ ≤ 1, -1 ≤ cos θ ≤ 1.

Key Points for Government Exams:

• Master the identities – they are used everywhere.
• Be quick with angle conversions (degree/radian).
• Understand the ASTC rule and allied angles thoroughly.
• Practice solving various types of trigonometric equations.
• Know the domain, range, and periodicity of all functions.
• Questions often involve finding max/min values or proving identities in MCQ format.
• Exemplar problems often test deeper understanding and application of multiple concepts.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts from Chapter 3, keeping the Exemplar style in mind:

1. If sin x + cosec x = 2, then sinⁿ x + cosecⁿ x is equal to:
   (A) 2
   (B) 2ⁿ
   (C) 2ⁿ⁻¹
   (D) 2ⁿ⁻²

2. The value of tan 1° tan 2° tan 3° ... tan 89° is:
   (A) 0
   (B) 1
   (C) ∞
   (D) 1/2

3. If tan θ = -4/3, and θ is in the second quadrant, then the value of sin θ + cos θ is:
   (A) 7/5
   (B) 1/5
   (C) -1/5
   (D) -7/5

4. The general solution of the equation tan 3x = tan 5x is:
   (A) x = nπ/2, n ∈ Z
   (B) x = nπ, n ∈ Z
   (C) x = nπ/8, n ∈ Z
   (D) x = nπ/(-2), n ∈ Z

5. The minimum value of 3 cos x + 4 sin x + 5 is:
   (A) 5
   (B) 9
   (C) 0
   (D) 1

6. If cos(α + β) = 4/5 and sin(α - β) = 5/13, where α, β lie between 0 and π/4, then tan 2α is equal to:
   (A) 19/12
   (B) 20/7
   (C) 25/16
   (D) 56/33

7. The number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π] is:
   (A) 0
   (B) 1
   (C) 2
   (D) 3

8. The value of cos (π/9) cos (2π/9) cos (4π/9) is:
   (A) 1/2
   (B) 1/4
   (C) 1/8
   (D) 1/16

9. An angle θ in degrees is such that tan θ = 7/24 and θ is in the third quadrant. The value of θ in radians is between:
   (A) π and 5π/4
   (B) 5π/4 and 3π/2
   (C) 3π/2 and 7π/4
   (D) 7π/4 and 2π

10. The domain of the function f(x) = √(1 - sin²x) + √(1 - cos²x) is:
    (A) R
    (B) [0, 2π]
    (C) R - {nπ | n ∈ Z}
    (D) R - {(2n+1)π/2 | n ∈ Z}

Answer Key:

1. (A)
2. (B)
3. (C)
4. (D) *Corrected: tan 5x - tan 3x = 0 => sin(5x-3x)/(cos5x cos3x) = 0 => sin 2x = 0 => 2x = nπ => x = nπ/2. However, need cos 3x ≠ 0 and cos 5x ≠ 0. If x=π/2, cos 3π/2=0. If x=π/6, cos 5π/6 ≠ 0, cos 3π/6 = cos π/2 = 0. Let's recheck the general solution tan A = tan B => A = nπ + B. So 5x = nπ + 3x => 2x = nπ => x = nπ/2. We must exclude values where cos 3x = 0 or cos 5x = 0. cos θ = 0 when θ = (2k+1)π/2. So 3x ≠ (2k+1)π/2 => x ≠ (2k+1)π/6. And 5x ≠ (2m+1)π/2 => x ≠ (2m+1)π/10. The solution x = nπ/2 includes odd multiples of π/2 (like π/2, 3π/2...) and integer multiples of π (like 0, π, 2π...). For x = (2p+1)π/2 (odd multiples), cos 3x = cos(3(2p+1)π/2) = 0 and cos 5x = cos(5(2p+1)π/2) = 0. So tan 3x and tan 5x are undefined. For x = pπ (integer multiples of π), cos 3x = cos 3pπ = (-1)³ᵖ ≠ 0 and cos 5x = cos 5pπ = (-1)⁵ᵖ ≠ 0. So tan 3x = 0 and tan 5x = 0. Thus tan 3x = tan 5x holds. The solution is x = pπ, n ∈ Z. So (B) is correct. Let me rethink the options provided. Maybe the question intended tan 5x = tan 3x. 5x = nπ + 3x => 2x = nπ => x = nπ/2. We need to exclude where cos 5x=0 or cos 3x=0. x != (2k+1)pi/10 and x != (2k+1)pi/6. x = nπ/2. If n is even, n=2p, x=pπ. cos(5pπ) != 0, cos(3pπ) != 0. So x=pπ is a solution. If n is odd, n=2p+1, x=(2p+1)π/2. cos(5(2p+1)π/2)=0, cos(3(2p+1)π/2)=0. So these are not solutions as tan is undefined. Therefore, the solution set is x = pπ, n ∈ Z. Option (B).
5. (C)
6. (D)
7. (D)
8. (C)
9. (B)
10. (A) *Corrected: √(1 - sin²x) = √cos²x = |cos x|. √(1 - cos²x) = √sin²x = |sin x|. f(x) = |cos x| + |sin x|. Both sin x and cos x are defined for all real numbers. So the domain is R. (A).

(Self-Correction during thought process for Q4 and Q10):

• For Q4, initially wrote down the formula correctly but didn't fully analyze the conditions for tan to be defined. Realized that solutions of the form (odd multiple) * π/2 make the denominators zero. The correct set of solutions is x = nπ. So option (B) is correct.
• For Q10, √(1-sin²x) simplifies to |cos x|, not just cos x. Similarly √(1-cos²x) = |sin x|. Since sin x and cos x are defined for all real numbers, their absolute values are also defined for all real numbers. Hence the domain is R. Option (A).

(Final Answer Key after correction):

1. (A)
2. (B)
3. (C)
4. (B)
5. (C)
6. (D)
7. (D)
8. (C)
9. (B)
10. (A)

Make sure you understand the steps and reasoning behind each concept and formula. Practice solving problems from the Exemplar book itself, as they often require careful application of these notes. Good luck!