Trigonometric Functions — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Trigonometric Functions" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Trigonometric Functions" — 8 important questions with detailed answers for CBSE bo…
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NCERT Solutions for Class 11 Mathematics — Trigonometric Functions. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Trigonometric Functions — All 8 Questions Solved
Q1. Find the value of sin(π/6), cos(π/4), and tan(π/3) using the unit circle or standard values.
Step 1: Recall standard trigonometric values for special angles.
For angle π/6 (30°):
sin(π/6) = 1/2
cos(π/6) = √3/2
tan(π/6) = 1/√3
For angle π/4 (45°):
sin(π/4) = 1/√2 = √2/2
cos(π/4) = 1/√2 = √2/2
tan(π/4) = 1
For angle π/3 (60°):
sin(π/3) = √3/2
cos(π/3) = 1/2
tan(π/3) = √3
Step 2: Identify required values.
sin(π/6) = 1/2
cos(π/4) = √2/2
tan(π/3) = √3
Final Answer: sin(π/6) = 1/2, cos(π/4) = √2/2, tan(π/3) = √3
Q2. Prove the trigonometric identity: (sin²θ + cos²θ) / cos²θ = sec²θ.
Step 1: Start with the left-hand side.
LHS = (sin²θ + cos²θ) / cos²θ
Step 2: Apply the Pythagorean identity.
We know that sin²θ + cos²θ = 1
Step 3: Substitute.
LHS = 1 / cos²θ
Step 4: Use the definition of secant.
sec θ = 1 / cos θ
Therefore, sec²θ = 1 / cos²θ
Step 5: Compare.
LHS = 1 / cos²θ = sec²θ = RHS ✓
Final Answer: The identity is proved: (sin²θ + cos²θ) / cos²θ = sec²θ
Q3. Find the general solution of the equation sin x = 1/2.
Step 1: Identify the principal solution.
We need sin x = 1/2
The principal value is x₀ = π/6 (or 30°)
Step 2: Recall the general solution for sin x = a.
If sin x = sin α, then:
x = nπ + (-1)ⁿ α, where n ∈ ℤ
Step 3: Apply the general solution formula.
For sin x = 1/2 = sin(π/6):
x = nπ + (-1)ⁿ (π/6), where n ∈ ℤ
Step 4: Expand for clarity.
When n is even (n = 2k):
x = 2kπ + π/6, k ∈ ℤ
When n is odd (n = 2k + 1):
x = (2k + 1)π - π/6 = 2kπ + π - π/6 = 2kπ + 5π/6, k ∈ ℤ
Final Answer: x = nπ + (-1)ⁿ(π/6), n ∈ ℤ
Or: x = 2kπ + π/6 or x = 2kπ + 5π/6, k ∈ ℤ
Q4. If sin θ = 3/5 and 0 < θ < π/2, find cos θ, tan θ, and sec θ.
Step 1: Given information.
sin θ = 3/5, and θ is in the first quadrant (0 < θ < π/2).
Step 2: Find cos θ using the Pythagorean identity.
sin²θ + cos²θ = 1
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 1 - 9/25 = 16/25
cos θ = ±4/5
Since θ is in the first quadrant, cos θ > 0.
cos θ = 4/5
Step 3: Find tan θ.
tan θ = sin θ / cos θ = (3/5) / (4/5) = 3/4
Step 4: Find sec θ.
sec θ = 1 / cos θ = 1 / (4/5) = 5/4
Step 5: Verify using another identity.
tan²θ + 1 = sec²θ
(3/4)² + 1 = (5/4)²
9/16 + 16/16 = 25/16
25/16 = 25/16 ✓
Final Answer: cos θ = 4/5, tan θ = 3/4, sec θ = 5/4
Q5. Prove that: cos(A + B) = cos A cos B - sin A sin B using the unit circle approach.
Step 1: Set up points on the unit circle.
Let P(A) = (cos A, sin A) be a point at angle A.
Let P(B) = (cos B, sin B) be a point at angle B.
Let P(A+B) = (cos(A+B), sin(A+B)) be a point at angle A+B.
Step 2: Consider the angle between P(A) and P(B).
The angle from P(B) to P(A) is A - B.
Step 3: Use the distance formula.
Distance from P(A) to P(B):
P(A)P(B)² = (cos A - cos B)² + (sin A - sin B)²
= cos²A - 2cos A cos B + cos²B + sin²A - 2sin A sin B + sin²B
= (cos²A + sin²A) + (cos²B + sin²B) - 2(cos A cos B + sin A sin B)
= 1 + 1 - 2(cos A cos B + sin A sin B)
= 2 - 2(cos A cos B + sin A sin B)
Step 4: Use the angle difference formula.
The angle between P(A) and P(B) is |A - B|.
Using the law of cosines in the unit circle:
P(A)P(B)² = 1 + 1 - 2(1)(1)cos(A - B)
= 2 - 2cos(A - B)
Step 5: Equate the two expressions.
2 - 2(cos A cos B + sin A sin B) = 2 - 2cos(A - B)
cos A cos B + sin A sin B = cos(A - B)
Step 6: Replace B with -B to get the addition formula.
cos(A + B) = cos(A - (-B)) = cos A cos(-B) + sin A sin(-B)
cos(A + B) = cos A cos B - sin A sin B
Final Answer: cos(A + B) = cos A cos B - sin A sin B is proved.
Q6. Find the period and amplitude of the function f(x) = 3 sin(2x) + 1.
Step 1: Identify the standard form.
The function is f(x) = 3 sin(2x) + 1
Compare with: f(x) = A sin(Bx + C) + D
Step 2: Extract parameters.
Amplitude |A| = |3| = 3
B = 2
C = 0 (phase shift = 0)
D = 1 (vertical shift)
Step 3: Find the period.
Period = 2π / |B| = 2π / 2 = π
Step 4: Find the range.
The basic sine function sin(2x) oscillates between -1 and 1.
Multiplying by 3: 3 sin(2x) oscillates between -3 and 3.
Adding 1: 3 sin(2x) + 1 oscillates between -3 + 1 = -2 and 3 + 1 = 4.
Range: [-2, 4]
Step 5: Verify with key points.
At x = 0: f(0) = 3 sin(0) + 1 = 1
At x = π/4: f(π/4) = 3 sin(π/2) + 1 = 3(1) + 1 = 4 (maximum)
At x = π/2: f(π/2) = 3 sin(π) + 1 = 1
At x = 3π/4: f(3π/4) = 3 sin(3π/2) + 1 = 3(-1) + 1 = -2 (minimum)
Final Answer: Amplitude = 3, Period = π, Range = [-2, 4]
Q7. Express cos 2θ in terms of sin θ, cos θ, or tan θ (three different forms).
Step 1: Use the double angle formula starting from cos(A + B) formula.
cos 2θ = cos(θ + θ) = cos θ cos θ - sin θ sin θ
cos 2θ = cos²θ - sin²θ ... (Form 1)
Step 2: Express in terms of cos θ only.
Using sin²θ = 1 - cos²θ:
cos 2θ = cos²θ - (1 - cos²θ)
cos 2θ = cos²θ - 1 + cos²θ
cos 2θ = 2cos²θ - 1 ... (Form 2)
Step 3: Express in terms of sin θ only.
Using cos²θ = 1 - sin²θ:
cos 2θ = (1 - sin²θ) - sin²θ
cos 2θ = 1 - 2sin²θ ... (Form 3)
Step 4: Express in terms of tan θ.
Using cos²θ - sin²θ = cos²θ(1 - tan²θ) and the identity 1/(1 + tan²θ) = cos²θ:
cos 2θ = (1 - tan²θ) / (1 + tan²θ) ... (Form 4)
Final Answer: cos 2θ has four equivalent forms:
(1) cos²θ - sin²θ
(2) 2cos²θ - 1
(3) 1 - 2sin²θ
(4) (1 - tan²θ) / (1 + tan²θ)
Q8. Solve the equation 2sin²x - sin x - 1 = 0 for 0 ≤ x < 2π.
Step 1: Let u = sin x and solve the quadratic.
2u² - u - 1 = 0
Step 2: Factor the quadratic.
2u² - 2u + u - 1 = 0
2u(u - 1) + 1(u - 1) = 0
(2u + 1)(u - 1) = 0
Step 3: Find values of u.
Either 2u + 1 = 0 or u - 1 = 0
u = -1/2 or u = 1
Step 4: Solve sin x = 1.
For 0 ≤ x < 2π:
sin x = 1 gives x = π/2
Step 5: Solve sin x = -1/2.
For 0 ≤ x < 2π:
The reference angle is π/6.
Since sin is negative in the 3rd and 4th quadrants:
x = π + π/6 = 7π/6 (3rd quadrant)
x = 2π - π/6 = 11π/6 (4th quadrant)
Step 6: Verify solutions.
For x = π/2: 2sin²(π/2) - sin(π/2) - 1 = 2(1) - 1 - 1 = 0 ✓
For x = 7π/6: 2(-1/2)² - (-1/2) - 1 = 2(1/4) + 1/2 - 1 = 1/2 + 1/2 - 1 = 0 ✓
For x = 11π/6: 2(-1/2)² - (-1/2) - 1 = 0 ✓
Final Answer: x = π/2, 7π/6, 11π/6
Trigonometric Functions — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The radian measure of 240° is:
- A. 2π/3
- B. 3π/4
- C. 4π/3
- D. 5π/3
Answer: C. 4π/3 — To convert degrees to radians, multiply the degree measure by π/180. So, 240° * (π/180°) = 4π/3 radians.
MCQ 2. If sin x < 0 and cos x < 0, then x lies in which quadrant?
- A. First quadrant
- B. Second quadrant
- C. Third quadrant
- D. Fourth quadrant
Answer: C. Third quadrant — In the third quadrant, both sine and cosine functions are negative. Tangent and cotangent are positive in this quadrant.
MCQ 3. If cos x = -3/5 and x lies in the third quadrant, then the value of sin x is:
- A. 4/5
- B. -4/5
- C. 3/5
- D. -3/5
Answer: B. -4/5 — We know sin²x + cos²x = 1. So, sin²x = 1 - (-3/5)² = 1 - 9/25 = 16/25. Thus, sin x = ±4/5. Since x is in the third quadrant, sin x is negative, so sin x = -4/5.
MCQ 4. The value of sin 75° is:
- A. (√3 + 1) / (2√2)
- B. (√3 - 1) / (2√2)
- C. 1 / (2√2)
- D. √3 / (2√2)
Answer: A. (√3 + 1) / (2√2) — We can write sin 75° as sin (45° + 30°). Using the formula sin(A+B) = sin A cos B + cos A sin B, we get (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1) / (2√2).
MCQ 5. The general solution of the equation sin x = √3/2 is:
- A. nπ ± π/3
- B. nπ + (-1)^n (π/3)
- C. 2nπ ± π/3
- D. 2nπ + π/3
Answer: B. nπ + (-1)^n (π/3) — For an equation of the form sin x = sin y, the general solution is x = nπ + (-1)^n y, where n ∈ Z. Since sin (π/3) = √3/2, the general solution is x = nπ + (-1)^n (π/3).
MCQ 6. The range of the function f(x) = sin x is:
- A. (-∞, ∞)
- B. [0, 1]
- C. [-1, 1]
- D. [-π/2, π/2]
Answer: C. [-1, 1] — The sine function oscillates between -1 and 1, inclusive. Therefore, its range is the closed interval [-1, 1].
MCQ 7. If cos A = 3/5, where A is in the first quadrant, then cos 2A is:
- A. 7/25
- B. -7/25
- C. 24/25
- D. -24/25
Answer: B. -7/25 — We use the double angle identity cos 2A = 2cos²A - 1. Substituting cos A = 3/5, we get cos 2A = 2(3/5)² - 1 = 2(9/25) - 1 = 18/25 - 1 = -7/25.
MCQ 8. The expression (sin 3x / sin x) - (cos 3x / cos x) simplifies to:
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — Combine the fractions: (sin 3x cos x - cos 3x sin x) / (sin x cos x). The numerator is sin(3x - x) = sin(2x). So, sin(2x) / (sin x cos x) = (2 sin x cos x) / (sin x cos x) = 2.
MCQ 9. The degree measure of 7π/6 radians is:
- A. 180°
- B. 210°
- C. 240°
- D. 270°
Answer: B. 210° — To convert radians to degrees, multiply the radian measure by 180°/π. So, (7π/6) * (180°/π) = 7 * 30° = 210°.
MCQ 10. The value of 2 sin 3x cos x is equivalent to:
- A. sin 4x + sin 2x
- B. cos 4x + cos 2x
- C. sin 4x - sin 2x
- D. cos 4x - cos 2x
Answer: A. sin 4x + sin 2x — Using the product-to-sum identity 2 sin A cos B = sin(A+B) + sin(A-B), we have 2 sin 3x cos x = sin(3x+x) + sin(3x-x) = sin 4x + sin 2x.
MCQ 11. If A+B+C = π, then sin 2A + sin 2B + sin 2C is equal to:
- A. 4 sin A sin B sin C
- B. 2 sin A sin B sin C
- C. cos A cos B cos C
- D. 0
Answer: A. 4 sin A sin B sin C — Using sum-to-product, sin 2A + sin 2B = 2 sin(A+B) cos(A-B). Since A+B = π-C, sin(A+B) = sin C. So, 2 sin C cos(A-B) + sin 2C = 2 sin C cos(A-B) + 2 sin C cos C = 2 sin C [cos(A-B) + cos C]. Since C = π-(A+B), cos C = -cos(A+B). So, 2 sin C [cos(A-B) - cos(A+B)] = 2 sin C [2 sin A sin B] = 4 sin A sin B sin C.
MCQ 12. The period of the function f(x) = sin(2x) is:
- A. π
- B. 2π
- C. π/2
- D. 4π
Answer: A. π — For a function of the form sin(ax+b), the period is 2π/|a|. Here, a=2, so the period is 2π/2 = π.
MCQ 13. The value of tan 15° is:
- A. 2 + √3
- B. 2 - √3
- C. 1 / (2 + √3)
- D. 1 / (2 - √3)
Answer: B. 2 - √3 — We can write tan 15° as tan(45° - 30°). Using tan(A-B) = (tan A - tan B) / (1 + tan A tan B), we get (1 - 1/√3) / (1 + 1/√3) = (√3 - 1) / (√3 + 1). Rationalizing gives 2 - √3.
MCQ 14. The number of solutions of tan x = 1 in the interval [0, 2π] is:
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — The principal value for tan x = 1 is x = π/4. In the interval [0, 2π], tangent is positive in the first and third quadrants. So, the solutions are x = π/4 and x = π + π/4 = 5π/4. There are exactly two solutions.
MCQ 15. The value of sin²x + cos²x is:
- A. 0
- B. 1
- C. 2
- D. -1
Answer: B. 1 — This is one of the fundamental trigonometric identities, true for all real values of x.
MCQ 16. If tan A = 1/2 and tan B = 1/3, then tan (A+B) is:
- A. 1
- B. 5/6
- C. 6/5
- D. 1/6
Answer: A. 1 — Using the formula tan(A+B) = (tan A + tan B) / (1 - tan A tan B), we have (1/2 + 1/3) / (1 - (1/2)(1/3)) = (5/6) / (1 - 1/6) = (5/6) / (5/6) = 1.
MCQ 17. The expression (sin 5x + sin 3x) / (cos 5x + cos 3x) is equal to:
- A. tan x
- B. tan 2x
- C. tan 3x
- D. tan 4x
Answer: D. tan 4x — Using sum-to-product identities: sin A + sin B = 2 sin((A+B)/2) cos((A-B)/2) and cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2). So, the expression becomes (2 sin(4x) cos(x)) / (2 cos(4x) cos(x)) = sin(4x) / cos(4x) = tan(4x).
MCQ 18. If an angle measures -11π/3 radians, in which quadrant does it terminate?
- A. First quadrant
- B. Second quadrant
- C. Third quadrant
- D. Fourth quadrant
Answer: A. First quadrant — To find the quadrant, add multiples of 2π until the angle is between 0 and 2π. -11π/3 + 4π = -11π/3 + 12π/3 = π/3. An angle of π/3 radians (60°) lies in the first quadrant.
MCQ 19. The general solution of tan x = tan (π/4) is:
- A. nπ + π/4
- B. 2nπ + π/4
- C. nπ ± π/4
- D. nπ + π/2
Answer: A. nπ + π/4 — For an equation of the form tan x = tan y, the general solution is x = nπ + y, where n ∈ Z. So, x = nπ + π/4.
MCQ 20. The maximum value of 3 sin x + 4 cos x is:
- A. 3
- B. 4
- C. 5
- D. 7
Answer: C. 5 — For an expression of the form a sin x + b cos x, the maximum value is √(a²+b²) and the minimum value is -√(a²+b²). Here, a=3, b=4, so the maximum value is √(3²+4²) = √25 = 5.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.