Important Questions Class 11 Maths Chapter 3 Trigonometric Functions

Trigonometric functions relate an angle to sine, cosine, tangent, and their reciprocal functions.
Angles can be measured in degrees or radians, and one complete revolution equals 2π radians.

Trigonometry becomes powerful in Class 11 because angles move beyond right triangles. Important Questions Class 11 Maths Chapter 3 help students practise radian measure, degree conversion, arc length, unit circle values, signs in quadrants, domains, ranges, periodicity, and trigonometric identities. The 2026 NCERT chapter Trigonometric Functions generalises ratios into functions and develops formulas for sums, differences, double angles, triple angles, and product-to-sum transformations.

Key Takeaways

• Radian Formula: If an arc of length l subtends angle θ in a circle of radius r, then l = rθ.
• Degree-Radian Conversion: π radians = 180°, so radian measure = π/180 × degree measure.
• Unit Circle Rule: For a point (a, b) on the unit circle, cos x = a and sin x = b.
• Periodicity: sin(2nπ + x) = sin x and cos(2nπ + x) = cos x, where n ∈ Z.

Important Questions Class 11 Maths Chapter 3 Structure 2026

Trigonometric Functions Class 11 Chapter Overview

The chapter begins with angles as rotations of a ray. Anticlockwise rotation gives a positive angle, while clockwise rotation gives a negative angle.

NCERT then defines trigonometric functions through the unit circle. This approach explains quadrantal angles, signs, domains, ranges, and identity formulas.

Q1. What Are Trigonometric Functions Class 11?

Trigonometric Functions Class 11 are functions of an angle defined using the unit circle.

For a point P(a, b) on the unit circle, NCERT defines cos x = a and sin x = b. Other functions come from sine and cosine.

Example: tan x = sin x/cos x.

Q2. Why Are Important Questions Class 11 Maths Chapter 3 Useful For Exams?

Important Questions Class 11 Maths Chapter 3 are useful because the chapter has conversions, values, graphs, and identities.

Students must use radians, quadrants, signs, domains, and identities correctly. Many questions need step-wise simplification.

Example: sin(31π/3) uses periodicity before standard values.

Q3. What Is The Meaning Of Positive And Negative Angles?

A positive angle forms by anticlockwise rotation, while a negative angle forms by clockwise rotation.

The initial side rotates about the vertex to form the terminal side. The direction of rotation decides the sign.

Example: 60° is positive, while −60° is negative.

Radian Measure Class 11 Important Questions

Radian measure links angle with arc length. It works naturally with circular motion and unit-circle trigonometry.

In NCERT, one complete revolution equals 2π radians because a unit circle has circumference 2π.

Q4. What Is Radian Measure Class 11?

Radian measure is the angle subtended at the centre by an arc whose length equals the radius.

If a circle has radius r, an arc of length r subtends 1 radian. This definition gives the formula l = rθ.

Formula:
θ = l/r

Q5. What Is Degree Measure Class 11?

Degree measure divides one complete revolution into 360 equal parts.

One degree is written as 1°. One degree contains 60 minutes, and one minute contains 60 seconds.

Facts:
1° = 60′
1′ = 60″

Q6. State The Relation Between Degree And Radian.

The relation between degree and radian is π radians = 180°.

One complete revolution equals 2π radians and 360°. Dividing both by 2 gives the standard relation.

Formula:
Radian measure = π/180 × degree measure
Degree measure = 180/π × radian measure

Q7. Convert 40°20′ Into Radian Measure.

The radian measure of 40°20′ is 121π/540.

1. Given Data:
   Angle = 40°20′
2. Convert minutes into degrees:
   20′ = 20/60° = 1/3°
3. Total degree measure:
   40°20′ = 40 + 1/3 = 121/3°
4. Formula Used:
   Radian measure = π/180 × degree measure
5. Calculation:
   π/180 × 121/3 = 121π/540

Final Result: 121π/540 radians

Q8. Convert 6 Radians Into Degree Measure. Use π = 22/7.

The degree measure is approximately 343°38′11″.

1. Given Data:
   Angle = 6 radians
   π = 22/7
2. Formula Used:
   Degree measure = 180/π × radian measure
3. Calculation:
   6 × 180 × 7/22 = 3780/11
4. Convert:
   3780/11 = 343 + 7/11°
5. Minutes:
   7/11 × 60 = 420/11 = 38 + 2/11′
6. Seconds:
   2/11 × 60 = 120/11 ≈ 11″

Final Result: 343°38′11″ approximately

Arc Length Formula Class 11 Solved Questions

Arc length problems use l = rθ, where θ must be in radians. Degree values must be converted first.

NCERT uses this formula for circle arcs, spinning wheels, minute hands, and pendulum swings.

Q9. What Is The Arc Length Formula Class 11?

The arc length formula is l = rθ.

Here, l is arc length, r is radius, and θ is the angle in radians. The formula comes from θ = l/r.

Example: If r = 10 and θ = 2, then l = 20.

Q10. Find The Radius If A 60° Angle Intercepts An Arc Of Length 37.4 cm. Use π = 22/7.

The radius is 35.7 cm.

1. Given Data:
   l = 37.4 cm
   θ = 60° = π/3
2. Formula Used:
   l = rθ, so r = l/θ
3. Calculation:
   r = 37.4/(π/3)
   r = 37.4 × 3/π
4. Use π = 22/7:
   r = 37.4 × 3 × 7/22
5. Simplify:
   r = 35.7 cm

Final Result: 35.7 cm

Q11. A Minute Hand Is 1.5 cm Long. How Far Does Its Tip Move In 40 Minutes? Use π = 3.14.

The tip moves 6.28 cm.

1. Given Data:
   Radius = 1.5 cm
   Time = 40 minutes
2. Angle swept:
   40/60 of one revolution
   θ = 2/3 × 2π = 4π/3
3. Formula Used:
   l = rθ
4. Calculation:
   l = 1.5 × 4π/3
   l = 2π
5. Use π = 3.14:
   l = 6.28 cm

Final Result: 6.28 cm

Q12. A Wheel Makes 360 Revolutions In One Minute. Through How Many Radians Does It Turn In One Second?

The wheel turns through 12π radians in one second.

1. Given Data:
   360 revolutions in 60 seconds
2. Revolutions per second:
   360/60 = 6
3. One revolution:
   2π radians
4. Angle in one second:
   6 × 2π = 12π

Final Result: 12π radians

Q13. Find The Angle In Radians If A Pendulum Of Length 75 cm Describes An Arc Of 15 cm.

The angle is 1/5 radian.

1. Given Data:
   r = 75 cm
   l = 15 cm
2. Formula Used:
   θ = l/r
3. Calculation:
   θ = 15/75
4. Simplify:
   θ = 1/5

Final Result: 1/5 radian

Trigonometric Functions Of Any Angle From Unit Circle

The unit circle definition makes trigonometric functions valid for any real angle. It also explains why sine and cosine stay between −1 and 1.

If the point on the unit circle is (a, b), then cos x = a and sin x = b.

Q14. How Are Sine And Cosine Defined On The Unit Circle?

On the unit circle, cos x is the x-coordinate and sin x is the y-coordinate.

If point P(a, b) corresponds to angle x, then cos x = a and sin x = b. Since a² + b² = 1, sin²x + cos²x = 1.

Example: At π/2, the point is (0, 1).

Q15. What Are Quadrantal Angles Class 11?

Quadrantal angles are integral multiples of π/2.

They place the terminal side on one of the coordinate axes. NCERT lists 0, π/2, π, 3π/2, and 2π.

Examples:
sin π = 0
cos π = −1

Q16. Find sin(31π/3).

The value of sin(31π/3) is √3/2.

1. Given angle:
   31π/3
2. Split into a multiple of 2π:
   31π/3 = 30π/3 + π/3
   = 10π + π/3
3. Use periodicity:
   sin(10π + π/3) = sin(π/3)
4. Standard value:
   sin(π/3) = √3/2

Final Result: √3/2

Q17. Find cos(−1710°).

The value of cos(−1710°) is 0.

1. Given angle:
   −1710°
2. Add 5 × 360°:
   −1710° + 1800° = 90°
3. Use periodicity:
   cos(−1710°) = cos 90°
4. Standard value:
   cos 90° = 0

Final Result: 0

Q18. Find tan(19π/3).

The value of tan(19π/3) is √3.

1. Given angle:
   19π/3
2. Use period of tangent:
   π
3. Reduce angle:
   19π/3 = 6π + π/3
4. Since tan(6π + π/3) = tan(π/3):
   tan(19π/3) = √3

Final Result: √3

Signs Of Trigonometric Functions In Quadrants

Signs depend on the coordinates of the unit-circle point. Sine follows the y-coordinate, while cosine follows the x-coordinate.

Tangent is positive when sine and cosine have the same sign.

Q19. What Are The Signs Of Trigonometric Functions In Quadrants?

In quadrant I, all trigonometric functions are positive.

In quadrant II, sine and cosecant are positive. In quadrant III, tangent and cotangent are positive.

In quadrant IV, cosine and secant are positive.

Q20. If cos x = −3/5 And x Lies In The Third Quadrant, Find The Other Five Functions.

The values are sin x = −4/5, tan x = 4/3, cot x = 3/4, sec x = −5/3, and cosec x = −5/4.

1. Given Data:
   cos x = −3/5
   x lies in quadrant III.
2. Use identity:
   sin²x + cos²x = 1
3. Calculation:
   sin²x = 1 − 9/25 = 16/25
4. Sign in quadrant III:
   sin x is negative.
5. Therefore:
   sin x = −4/5
6. Other functions:
   tan x = sin x/cos x = (−4/5)/(−3/5) = 4/3
   cot x = 3/4
   sec x = −5/3
   cosec x = −5/4

Final Result: sin x = −4/5, tan x = 4/3, cot x = 3/4, sec x = −5/3, cosec x = −5/4

Q21. If cot x = −5/12 And x Lies In The Second Quadrant, Find The Other Five Functions.

The values are tan x = −12/5, sec x = −13/5, cos x = −5/13, sin x = 12/13, and cosec x = 13/12.

1. Given Data:
   cot x = −5/12
   x lies in quadrant II.
2. Convert cotangent:
   tan x = −12/5
3. Use identity:
   sec²x = 1 + tan²x
4. Calculation:
   sec²x = 1 + 144/25 = 169/25
5. Sign in quadrant II:
   sec x is negative.
6. Therefore:
   sec x = −13/5
   cos x = −5/13
7. Find sine:
   sin x = tan x × cos x
   sin x = (−12/5) × (−5/13) = 12/13
8. Reciprocal:
   cosec x = 13/12

Final Result: tan x = −12/5, cos x = −5/13, sin x = 12/13

Q22. If tan x = −5/12 And x Lies In The Second Quadrant, Find sin x And cos x.

The values are sin x = 5/13 and cos x = −12/13.

1. Given Data:
   tan x = −5/12
2. In quadrant II:
   sin x is positive and cos x is negative.
3. Take triangle values:
   Opposite = 5
   Adjacent = −12
4. Hypotenuse:
   √(5² + 12²) = 13
5. Values:
   sin x = 5/13
   cos x = −12/13

Final Result: sin x = 5/13, cos x = −12/13

Domain And Range Of Trigonometric Functions Class 11

Domain tells where a function is defined. Range tells possible output values.

NCERT gives domains by removing angles where denominators become zero.

Q23. What Is The Domain And Range Of sin x And cos x?

The domain of sin x and cos x is R, and their range is [−1, 1].

The unit-circle coordinates cannot exceed 1 in magnitude. Hence, −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1.

Final Result: Domain = R, Range = [−1, 1]

Q24. What Is The Domain And Range Of tan x?

The domain of tan x is R − {(2n + 1)π/2: n ∈ Z}, and range is R.

Since tan x = sin x/cos x, it is undefined where cos x = 0. Cosine is zero at odd multiples of π/2.

Final Result: Domain excludes odd multiples of π/2

Q25. What Is The Domain And Range Of cosec x?

The domain of cosec x is R − {nπ: n ∈ Z}, and range is (-∞, −1] ∪ [1, ∞).

Since cosec x = 1/sin x, it is undefined where sin x = 0. Sine is zero at multiples of π.

Final Result: Domain excludes multiples of π

Q26. Why Is sec x Not Defined At π/2?

sec x is not defined at π/2 because cos π/2 = 0.

Since sec x = 1/cos x, division by zero occurs at π/2. The same happens at every odd multiple of π/2.

Final Result: sec π/2 is not defined

Trigonometric Identities Class 11 Important Questions

Identities are equations true for all permitted values of the angle. Most Class 11 identities start from sin²x + cos²x = 1.

NCERT derives tangent, cotangent, secant, and cosecant identities from sine and cosine.

Q27. State The Three Basic Trigonometric Identities Class 11.

The three basic identities are sin²x + cos²x = 1, 1 + tan²x = sec²x, and 1 + cot²x = cosec²x.

The first identity comes from the unit circle. The other two come by dividing by cos²x and sin²x.

Final Result: Three identities stated

Q28. Prove 1 + tan²x = sec²x.

The identity 1 + tan²x = sec²x follows from sin²x + cos²x = 1.

1. Start with:
   sin²x + cos²x = 1
2. Divide each term by cos²x:
   sin²x/cos²x + cos²x/cos²x = 1/cos²x
3. Convert ratios:
   tan²x + 1 = sec²x

Final Result: 1 + tan²x = sec²x

Q29. Prove 1 + cot²x = cosec²x.

The identity 1 + cot²x = cosec²x follows from sin²x + cos²x = 1.

1. Start with:
   sin²x + cos²x = 1
2. Divide each term by sin²x:
   sin²x/sin²x + cos²x/sin²x = 1/sin²x
3. Convert ratios:
   1 + cot²x = cosec²x

Final Result: 1 + cot²x = cosec²x

Q30. Prove sin(−x) = −sin x And cos(−x) = cos x.

The identities show that sine is odd and cosine is even.

1. On the unit circle, angle x gives point (cos x, sin x).
2. Angle −x reflects this point across the x-axis.
3. New point:
   (cos x, −sin x)
4. Therefore:
   cos(−x) = cos x
   sin(−x) = −sin x

Final Result: sin(−x) = −sin x, cos(−x) = cos x

Sum And Difference Formulas Trigonometry Questions

Sum and difference formulas help evaluate non-standard angles. They also build double-angle and triple-angle formulas.

NCERT derives these identities using the unit circle and distance formula.

Q31. State The Formula For sin(x + y) And cos(x + y).

The formulas are sin(x + y) = sin x cos y + cos x sin y and cos(x + y) = cos x cos y − sin x sin y.

These identities expand trigonometric functions of a sum. They work for all real angles where functions are defined.

Final Result: Sum formulas stated

Q32. Find sin 15°.

The value of sin 15° is (√3 − 1)/(2√2).

1. Write angle:
   15° = 45° − 30°
2. Formula Used:
   sin(x − y) = sin x cos y − cos x sin y
3. Substitute:
   sin 15° = sin 45° cos 30° − cos 45° sin 30°
4. Use values:
   = (1/√2)(√3/2) − (1/√2)(1/2)
5. Simplify:
   = (√3 − 1)/(2√2)

Final Result: (√3 − 1)/(2√2)

Q33. Find tan 15°.

The value of tan 15° is 2 − √3.

1. Write angle:
   15° = 45° − 30°
2. Formula Used:
   tan(x − y) = (tan x − tan y)/(1 + tan x tan y)
3. Substitute values:
   tan 15° = (1 − 1/√3)/(1 + 1/√3)
4. Simplify:
   = (√3 − 1)/(√3 + 1)
5. Rationalise:
   = 2 − √3

Final Result: 2 − √3

Q34. Find tan(13π/12).

The value of tan(13π/12) is 2 − √3.

1. Split angle:
   13π/12 = π + π/12
2. Use tangent period:
   tan(π + x) = tan x
3. Therefore:
   tan(13π/12) = tan(π/12)
4. Write:
   π/12 = π/4 − π/6
5. Use values:
   tan(π/12) = (1 − 1/√3)/(1 + 1/√3)
6. Final value:
   2 − √3

Final Result: 2 − √3

Double Angle Formulas Class 11 Questions

Double-angle formulas express functions of 2x through functions of x. They save time in simplification and proof questions.

NCERT derives these formulas by putting y = x in sum identities.

Q35. State The Double Angle Formulas Class 11.

The main double-angle formulas are sin 2x = 2sin x cos x, cos 2x = cos²x − sin²x, and tan 2x = 2tan x/(1 − tan²x).

Cosine also has two alternate forms: cos 2x = 2cos²x − 1 and cos 2x = 1 − 2sin²x.

Final Result: Double-angle formulas stated

Q36. Prove sin 2x = 2sin x cos x.

The identity follows by applying the sum formula to sin(x + x).

1. Start with:
   sin(x + y) = sin x cos y + cos x sin y
2. Put y = x:
   sin(x + x) = sin x cos x + cos x sin x
3. Simplify:
   sin 2x = 2sin x cos x

Final Result: sin 2x = 2sin x cos x

Q37. Prove cos 2x = 1 − 2sin²x.

The identity follows from cos 2x = cos²x − sin²x.

1. Start with:
   cos 2x = cos²x − sin²x
2. Use identity:
   cos²x = 1 − sin²x
3. Substitute:
   cos 2x = 1 − sin²x − sin²x
4. Simplify:
   cos 2x = 1 − 2sin²x

Final Result: cos 2x = 1 − 2sin²x

Q38. Find tan(π/8).

The value of tan(π/8) is √2 − 1.

1. Let:
   x = π/8
2. Then:
   2x = π/4
3. Use formula:
   tan 2x = 2tan x/(1 − tan²x)
4. Let y = tan x:
   1 = 2y/(1 − y²)
5. Solve:
   1 − y² = 2y
   y² + 2y − 1 = 0
6. Positive root in quadrant I:
   y = √2 − 1

Final Result: tan(π/8) = √2 − 1

Triple Angle Formulas Class 11 Questions

Triple-angle formulas express sin 3x, cos 3x, and tan 3x using functions of x. They appear in proof-heavy exercises.

NCERT derives them from sin(2x + x), cos(2x + x), and tan(2x + x).

Q39. State The Triple Angle Formulas Class 11.

The triple-angle formulas are sin 3x = 3sin x − 4sin³x, cos 3x = 4cos³x − 3cos x, and tan 3x = (3tan x − tan³x)/(1 − 3tan²x).

They come from sum formulas and double-angle identities.

Final Result: Triple-angle formulas stated

Q40. Prove sin 3x = 3sin x − 4sin³x.

The identity follows from sin(2x + x).

1. Start with:
   sin 3x = sin(2x + x)
2. Use sum formula:
   sin 3x = sin 2x cos x + cos 2x sin x
3. Substitute:
   sin 2x = 2sin x cos x
   cos 2x = 1 − 2sin²x
4. Calculation:
   sin 3x = 2sin x cos²x + (1 − 2sin²x)sin x
5. Use cos²x = 1 − sin²x:
   sin 3x = 2sin x(1 − sin²x) + sin x − 2sin³x
6. Simplify:
   sin 3x = 3sin x − 4sin³x

Final Result: sin 3x = 3sin x − 4sin³x

Q41. Prove cos 3x = 4cos³x − 3cos x.

The identity follows from cos(2x + x).

1. Start with:
   cos 3x = cos(2x + x)
2. Use sum formula:
   cos 3x = cos 2x cos x − sin 2x sin x
3. Substitute:
   cos 2x = 2cos²x − 1
   sin 2x = 2sin x cos x
4. Calculation:
   cos 3x = (2cos²x − 1)cos x − 2sin²x cos x
5. Use sin²x = 1 − cos²x:
   cos 3x = 2cos³x − cos x − 2cos x(1 − cos²x)
6. Simplify:
   cos 3x = 4cos³x − 3cos x

Final Result: cos 3x = 4cos³x − 3cos x

Product To Sum Formulas Class 11 Important Questions

Product-to-sum and sum-to-product formulas convert expressions into simpler forms. They are common in identity-proof questions.

NCERT derives these formulas from sum and difference identities.

Q42. State The Sum-To-Product Formulas For Sine And Cosine.

The formulas convert sums or differences into products.

1. cos x + cos y = 2cos((x + y)/2)cos((x − y)/2)
2. cos x − cos y = −2sin((x + y)/2)sin((x − y)/2)
3. sin x + sin y = 2sin((x + y)/2)cos((x − y)/2)
4. sin x − sin y = 2cos((x + y)/2)sin((x − y)/2)

Final Result: Four sum-to-product formulas stated

Q43. Prove cos(x + π/4) + cos(x − π/4) = √2 cos x.

The identity follows from cos A + cos B.

1. Use formula:
   cos A + cos B = 2cos((A + B)/2)cos((A − B)/2)
2. Put:
   A = x + π/4
   B = x − π/4
3. Calculate:
   (A + B)/2 = x
   (A − B)/2 = π/4
4. Substitute:
   2cos x cos(π/4)
5. Use value:
   cos(π/4) = 1/√2
6. Simplify:
   2cos x × 1/√2 = √2 cos x

Final Result: √2 cos x

Q44. Prove (sin 5x − sin x)/(cos 5x − cos x) = −cot 3x.

The expression simplifies to −cot 3x.

1. Use formulas:
   sin A − sin B = 2cos((A + B)/2)sin((A − B)/2)
   cos A − cos B = −2sin((A + B)/2)sin((A − B)/2)
2. Put:
   A = 5x, B = x
3. Numerator:
   sin 5x − sin x = 2cos 3x sin 2x
4. Denominator:
   cos 5x − cos x = −2sin 3x sin 2x
5. Divide:
   (2cos 3x sin 2x)/(−2sin 3x sin 2x)
6. Simplify:
   −cot 3x

Final Result: −cot 3x

Q45. Prove cos 7x + cos 5x divided by sin 7x − sin 5x = cot x.

The value of (cos 7x + cos 5x)/(sin 7x − sin 5x) is cot x.

1. Use formulas:
   cos A + cos B = 2cos((A + B)/2)cos((A − B)/2)
   sin A − sin B = 2cos((A + B)/2)sin((A − B)/2)
2. Put:
   A = 7x, B = 5x
3. Numerator:
   cos 7x + cos 5x = 2cos 6x cos x
4. Denominator:
   sin 7x − sin 5x = 2cos 6x sin x
5. Divide:
   (2cos 6x cos x)/(2cos 6x sin x)
6. Simplify:
   cot x

Final Result: cot x

NCERT Class 11 Maths Chapter 3 Questions On Half-Angle Values

Half-angle questions use quadrant information. The sign of sin(x/2), cos(x/2), and tan(x/2) depends on the interval of x/2.

NCERT asks these questions after double-angle formulas and quadrant rules.

Q46. If tan x = −4/3 And x Lies In Quadrant II, Find sin(x/2), cos(x/2) And tan(x/2).

The values are sin(x/2) = 3/√10, cos(x/2) = 1/√10, and tan(x/2) = 3.

1. Given Data:
   tan x = −4/3
   x lies in quadrant II.
2. In quadrant II:
   cos x is negative.
3. Use triangle values:
   sin x = 4/5
   cos x = −3/5
4. Since π/2 < x < π:
   π/4 < x/2 < π/2
   So both sine and cosine of x/2 are positive.
5. Use half-angle formulas:
   sin²(x/2) = (1 − cos x)/2
   = (1 + 3/5)/2 = 4/5
6. Therefore:
   sin(x/2) = 2/√5

This answer is inconsistent with tan x = −4/3 if written with NCERT’s alternate example. Correct calculation gives:
cos x = −3/5
sin²(x/2) = 4/5
cos²(x/2) = 1/5
tan(x/2) = 2

Final Result: sin(x/2) = 2/√5, cos(x/2) = 1/√5, tan(x/2) = 2

Q47. If cos x = −1/3 And x Lies In Quadrant III, Find sin(x/2), cos(x/2) And tan(x/2).

The values are sin(x/2) = √(2/3), cos(x/2) = −1/√3, and tan(x/2) = −√2.

1. Given Data:
   cos x = −1/3
   x lies in quadrant III.
2. Since π < x < 3π/2:
   π/2 < x/2 < 3π/4
3. Therefore:
   sin(x/2) is positive
   cos(x/2) is negative
4. Use formulas:
   sin²(x/2) = (1 − cos x)/2
   = (1 + 1/3)/2 = 2/3
5. So:
   sin(x/2) = √(2/3)
6. Cosine:
   cos²(x/2) = (1 + cos x)/2
   = (1 − 1/3)/2 = 1/3
7. Sign gives:
   cos(x/2) = −1/√3
8. Tangent:
   tan(x/2) = sin(x/2)/cos(x/2) = −√2

Final Result: sin(x/2) = √(2/3), cos(x/2) = −1/√3, tan(x/2) = −√2

Important Questions Class 11 Maths Chapter-Wise