Trigonometry — Exercise Solutions
1. How to Use This Page
This page covers all exercise types from Chapter 3 (Trigonometry) of the TN State Board Class 11 Maths textbook. Work through each section in order:
- MCQs — test conceptual clarity and quick recall
- Fill in the Blanks — reinforce formulas and identities
- True or False — catch common misconceptions
- Match the Following — connect formulas to their names or results
- Brief Answers — 2-mark style proofs and derivations
- Detailed Answers — 5-mark style worked numerical examples
Read the Key Concepts recap first if you need a refresher, then attempt the questions on your own before reading the solutions.
2. Key Concepts (Quick Recap)
2.1 Radian Measure
| Concept | Formula |
|---|---|
| Radian definition | Angle subtended by an arc equal in length to the radius |
| Degree → Radian | θ rad = θ° × π / 180 |
| Radian → Degree | θ° = θ rad × 180 / π |
| Arc length | l = rθ (θ in radians) |
| Full circle | 2π rad = 360° |
2.2 Trigonometric Functions — Signs by Quadrant
| Quadrant | sin | cos | tan | cosec | sec | cot |
|---|---|---|---|---|---|---|
| I (0°–90°) | + | + | + | + | + | + |
| II (90°–180°) | + | − | − | + | − | − |
| III (180°–270°) | − | − | + | − | − | + |
| IV (270°–360°) | − | + | − | − | + | − |
Mnemonic: All Students Take Coffee — All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4.
2.3 Pythagorean Identities
| Identity | Alternate Form |
|---|---|
| sin²θ + cos²θ = 1 | sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ |
| 1 + tan²θ = sec²θ | tan²θ = sec²θ − 1 |
| 1 + cot²θ = cosec²θ | cot²θ = cosec²θ − 1 |
2.4 Compound Angle Formulas
| Formula | Expression |
|---|---|
| sin(A + B) | sin A cos B + cos A sin B |
| sin(A − B) | sin A cos B − cos A sin B |
| cos(A + B) | cos A cos B − sin A sin B |
| cos(A − B) | cos A cos B + sin 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) |
2.5 Double Angle Formulas
| Formula | Expression |
|---|---|
| sin 2θ | 2 sin θ cos θ |
| cos 2θ | cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ |
| tan 2θ | 2tan θ / (1 − tan²θ) |
2.6 Transformation Formulas
Product-to-Sum:
- 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)
Sum-to-Product:
- 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)
2.7 General Solutions of Trigonometric Equations
| Equation | General Solution |
|---|---|
| sin θ = sin α | θ = nπ + (−1)ⁿα, n ∈ ℤ |
| cos θ = cos α | θ = 2nπ ± α, n ∈ ℤ |
| tan θ = tan α | θ = nπ + α, n ∈ ℤ |
| sin θ = 0 | θ = nπ, n ∈ ℤ |
| cos θ = 0 | θ = (2n + 1)π/2, n ∈ ℤ |
| tan θ = 0 | θ = nπ, n ∈ ℤ |
3. I. Choose the Best Answer (MCQs)
Q1. The value of sin(π/6) + cos(π/3) is
(a) 1/2 (b) 1 (c) √3/2 (d) 0
Q2. If sin θ = 3/5 and θ is in the second quadrant, then cos θ equals
(a) 4/5 (b) −4/5 (c) −3/5 (d) 3/5
Q3. The general solution of cos θ = 1/2 is
(a) θ = 2nπ ± π/6 (b) θ = 2nπ ± π/3 (c) θ = nπ ± π/3 (d) θ = nπ + (−1)ⁿ(π/3)
Q4. sin 75° equals
(a) (√6 + √2)/4 (b) (√6 − √2)/4 (c) (√3 + 1)/4 (d) (√3 − 1)/4
Q5. The value of 2 sin 45° cos 15° is
(a) sin 60° + sin 30° (b) sin 60° − sin 30° (c) cos 60° + cos 30° (d) cos 60° − cos 30°
Q6. cos²15° − sin²15° equals
(a) 1/2 (b) √3/2 (c) √2/2 (d) 1
4. II. Fill in the Blanks
1. The angle subtended at the centre of a circle by an arc equal in length to the radius is ______. 1 radian
2. 270° = ______ radians. 3π/2 Explanation: 270 × π/180 = 3π/2
3. If sin θ = 4/5 and θ ∈ (π/2, π), then tan θ = ______. −4/3 Explanation: cos θ = −3/5 (Q2), so tan θ = sin θ/cos θ = (4/5)/(−3/5) = −4/3
4. The range of the function f(θ) = 3 sin θ + 4 cos θ is ______. [−5, 5] Explanation: a sin θ + b cos θ has range [−√(a²+b²), √(a²+b²)] = [−5, 5]
5. sin(A + B) sin(A − B) = ______. sin²A − sin²B Explanation: Use product-to-sum or expand both sides
6. The general solution of tan θ = √3 is θ = ______. nπ + π/3, n ∈ ℤ
7. cos(A − B) − cos(A + B) = ______. 2 sin A sin B
8. If cos 2θ = 1/2, then cos θ = ______. ±√3/2 Explanation: 2cos²θ − 1 = 1/2 → cos²θ = 3/4 → cos θ = ±√3/2
9. The value of sin(π/3) cos(π/6) + cos(π/3) sin(π/6) is ______. 1 Explanation: This is sin(π/3 + π/6) = sin(π/2) = 1
10. The period of the function f(x) = tan 3x is ______. π/3 Explanation: Period of tan kx = π/k, so period = π/3
11. 2 sin 15° sin 75° = ______. 1/2 Explanation: 2 sin 15° sin 75° = cos(15° − 75°) − cos(15° + 75°) = cos 60° − cos 90° = 1/2 − 0 = 1/2
12. The minimum value of 5 sin θ − 12 cos θ + 7 is ______. −6 Explanation: 5 sin θ − 12 cos θ has range [−13, 13]. Minimum of the expression = −13 + 7 = −6.
5. III. True or False (with corrections)
Q1. "sin²θ + cos²θ = 2" — True or False?
False. The correct identity is sin²θ + cos²θ = 1.
Q2. "The value of sin θ can be greater than 1" — True or False?
False. For all real θ, −1 ≤ sin θ ≤ 1. The range of sin θ is [−1, 1].
Q3. "cos(A + B) = cos A cos B + sin A sin B" — True or False?
False. The correct formula is cos(A + B) = cos A cos B − sin A sin B. The plus sign between the terms is for cos(A − B).
Q4. "The general solution of sin θ = 0 is θ = nπ, n ∈ ℤ" — True or False?
True. The sine function equals zero at integer multiples of π: θ = nπ, n ∈ ℤ.
Q5. "tan θ is defined for all real values of θ" — True or False?
False. tan θ is undefined at θ = (2n + 1)π/2, n ∈ ℤ, where cos θ = 0. The domain of tan θ excludes these points.
Q6. "If sin θ = −1/2 and θ is in the third quadrant, then θ = 7π/6" — True or False?
True. In Q3, sin θ is negative. sin(7π/6) = sin(π + π/6) = −sin(π/6) = −1/2.
6. IV. Match the Following
| Column A | Column B |
|---|---|
| (A) sin(A + B) | (1) cos A cos B − sin A sin B |
| (B) cos(A − B) | (2) (tan A − tan B)/(1 + tan A tan B) |
| (C) tan(A + B) | (3) sin A cos B + cos A sin B |
| (D) 2 sin A cos B | (4) 2 cos((C+D)/2) cos((C−D)/2) |
| (E) cos C + cos D | (5) sin(A + B) + sin(A − B) |
| (F) tan(A − B) | (6) cos A cos B + sin A sin B |
- (A) → (3): sin(A + B) = sin A cos B + cos A sin B
- (B) → (6): cos(A − B) = cos A cos B + sin A sin B
- (C): tan(A + B) = (tan A + tan B)/(1 − tan A tan B) — not directly in Column B
- (D) → (5): 2 sin A cos B = sin(A + B) + sin(A − B)
- (E) → (4): cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2)
- (F) → (2): tan(A − B) = (tan A − tan B)/(1 + tan A tan B)
7. V. Answer Briefly (2-Mark Questions)
Q1. Prove that sin(π − θ) = sin θ for all θ.
Q2. Convert 150° into radians and find sin 150°.
Q3. Express cos 2θ in three different forms.
The three forms of cos 2θ are:
1. cos 2θ = cos²θ − sin²θ 2. cos 2θ = 2cos²θ − 1 3. cos 2θ = 1 − 2sin²θ
Q4. Find the value of tan(π/12).
Q5. Prove that (sin θ + cos θ)² + (sin θ − cos θ)² = 2.
8. VI. Answer in Detail (5-Mark Questions)
Q1. Prove the compound angle formula for sin(A + B) and use it to find sin 75°.
Proof of sin(A + B):
Consider a unit circle. Let OP make angle A with the x-axis, and OQ make angle (A + B) with the x-axis.
Q2. Derive the double angle formulas from compound angle formulas and solve: If cos θ = 3/5, find sin 2θ, cos 2θ, and tan 2θ when θ is in the first quadrant.
Derivation from compound angle formulas:
Numerical Solution:
Q3. Find the general solution of the equation 2cos²θ + 3sin θ = 3.
Q4. Prove the product-to-sum identity: 2 sin A cos B = sin(A + B) + sin(A − B). Hence express sin 50° cos 20° as a sum.
Proof:
Application:
9. Common Mistakes to Avoid
1. Forgetting quadrant signs. Always determine which quadrant the angle lies in before assigning signs. A common error is writing cos θ = 4/5 without the negative sign when θ is in Q2.
2. Mixing up cos(A + B) and cos(A − B). The formulas differ only in the sign between the terms: cos(A + B) = cos A cos B − sin A sin B, but cos(A − B) = cos A cos B + sin A sin B. Note the swapped sign.
3. Incorrect general solutions. Do not confuse the patterns:
Each has a distinct structure; mixing them up costs marks.
- sin θ = sin α → θ = nπ + (−1)ⁿα
- cos θ = cos α → θ = 2nπ ± α
- tan θ = tan α → θ = nπ + α
4. Radian-degree confusion. When the argument of a trig function is in radians, do not multiply by π/180 again. Remember: sin(π/3) means the angle is already in radians (60°). Only convert when the angle is given in degrees.
5. Squaring introduces extraneous solutions. When solving trig equations, if you square both sides (e.g., going from sin θ = cos θ to sin²θ = cos²θ), always verify each solution in the original equation. Squaring can introduce false roots.
6. Domain restrictions on tan θ. Remember that tan θ is undefined at odd multiples of π/2. When writing general solutions involving tan, exclude these values implicitly — but do not divide by an expression that could be zero.
7. Forgetting the ± in square root identities. When using cos²θ = 1 − sin²θ, remember cos θ = ±√(1 − sin²θ). The sign depends on the quadrant. Blindly taking the positive root is a frequent error.
10. Exam Checklist
Before entering the exam hall, confirm you can do each of the following without notes:
- [ ] Convert between degrees and radians fluently (e.g., 135° = 3π/4 rad, π/5 = 36°)
- [ ] State all six trig function values at standard angles: 0, π/6, π/4, π/3, π/2, π
- [ ] Recite the three Pythagorean identities from memory
- [ ] Write out all six compound angle formulas (sin, cos, tan — both + and −)
- [ ] Derive the three double angle formulas from the compound angle formulas
- [ ] Apply product-to-sum and sum-to-product identities in both directions
- [ ] Determine the sign of every trig function in each of the four quadrants (ASTC rule)
- [ ] Solve sin θ = k, cos θ = k, and tan θ = k for general solutions
- [ ] Handle quadratic trig equations by substitution (e.g., 2sin²θ + 3sin θ − 2 = 0)
- [ ] Find the range of a sin θ + b cos θ as [−√(a²+b²), √(a²+b²)]
- [ ] Rationalise results involving √2, √3 to get standard surd forms like (√6 ± √2)/4
- [ ] State the domain and range, and the period, of all six trig functions
Time management tip: In a 3-hour exam, spend no more than 2 minutes per MCQ and 8–10 minutes per 5-mark question. If a derivation question is taking longer, move on and return to it later.