JEE Main Mathematics · Chapter 19

Trigonometry

Identities, trigonometric equations, inverse trigonometric functions and properties of triangles.

NCERT Class 11 & 12JEE Main: 1–2 Qs per shift 7 min read

1. Chapter at a Glance

Trigonometry studies the relations between angles and lengths. In JEE it has four parts: trigonometric ratios and identities, trigonometric equations, inverse trigonometric functions and properties of triangles. Trigonometry also runs through calculus, complex numbers and coordinate geometry, so fluency with identities is essential. JEE Main asks one or two direct questions.

What JEE asks most
  • Number of solutions of a trigonometric equation in a given interval.
  • Maximum and minimum values of trigonometric expressions.
  • Simplifying inverse trigonometric expressions using principal values.
  • Products like cos 20° cos 40° cos 80° and sums using standard identities.

2. Trigonometric Ratios and the Unit Circle

On the unit circle, a point at angle θ from the positive x-axis has coordinates (cos θ, sin θ); tan θ = sin θ/cos θ. The signs of the ratios in the four quadrants follow "All Students Take Coffee": All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4.

(cos θ, sin θ) θ I: All +II: Sin + III: Tan +IV: Cos +
Unit circle: x = cos θ, y = sin θ. The quadrant labels show which ratios are positive.
θ0°30°45°60°90°
sin θ01/21/√2√3/21
cos θ1√3/21/√21/20
tan θ01/√31√3not defined

Useful values: sin 15° = (√6 − √2)/4, cos 15° = (√6 + √2)/4, tan 15° = 2 − √3, sin 18° = (√5 − 1)/4, cos 36° = (√5 + 1)/4.

3. Identities

  • sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ.
  • sin(A ± B) = sin A cos B ± cos 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).
  • sin 2A = 2 sin A cos A = 2tan A/(1 + tan²A); cos 2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1.
  • sin 3A = 3 sin A − 4 sin³A; cos 3A = 4 cos³A − 3 cos A; tan 3A = (3tan A − tan³A)/(1 − 3tan²A).
  • Sum to product: sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2); cos C + cos D = 2 cos((C + D)/2) cos((C − D)/2).
  • Product to sum: 2 sin A cos B = sin(A + B) + sin(A − B); 2 cos A cos B = cos(A + B) + cos(A − B).
Standard results
  • a sin θ + b cos θ lies between −√(a² + b²) and √(a² + b²).
  • sin θ sin(60° − θ) sin(60° + θ) = (1/4) sin 3θ; cos θ cos(60° − θ) cos(60° + θ) = (1/4) cos 3θ.
  • cos A cos 2A cos 4A … cos 2ⁿ⁻¹A = sin 2ⁿA/(2ⁿ sin A).

4. Trigonometric Equations

EquationGeneral solution (n ∈ ℤ)
sin θ = sin αθ = nπ + (−1)ⁿα
cos θ = cos αθ = 2nπ ± α
tan θ = tan αθ = nπ + α
sin²θ = sin²α (or cos², tan²)θ = nπ ± α
sin θ = 0 / cos θ = 0θ = nπ / θ = (2n + 1)π/2

To count solutions in an interval, factorise, solve each factor separately, list the solutions in the interval, and remove any repeats or values where the equation is undefined.

5. Inverse Trigonometric Functions

FunctionDomainPrincipal value range
sin−1x[−1, 1][−π/2, π/2]
cos−1x[−1, 1][0, π]
tan−1xℝ(−π/2, π/2)
cot−1xℝ(0, π)
sec−1x|x| ≥ 1[0, π] − {π/2}
cosec−1x|x| ≥ 1[−π/2, π/2] − {0}
  • sin−1x + cos−1x = π/2; tan−1x + cot−1x = π/2.
  • sin−1(−x) = −sin−1x, tan−1(−x) = −tan−1x, but cos−1(−x) = π − cos−1x.
  • tan−1x + tan−1y = tan−1((x + y)/(1 − xy)) when xy < 1.
  • sin−1(sin θ) = θ only when θ ∈ [−π/2, π/2]; otherwise bring θ into the principal range first.

6. Properties of Triangles

  • Sine rule: a/sin A = b/sin B = c/sin C = 2R (R = circumradius).
  • Cosine rule: cos A = (b² + c² − a²)/(2bc).
  • Area: Δ = ½ bc sin A = √(s(s − a)(s − b)(s − c)) (Heron), where s = (a + b + c)/2.
  • R = abc/(4Δ); inradius r = Δ/s.
  • Projection formula: a = b cos C + c cos B.

7. Formula Master Table

ResultFormula
Pythagoreansin²θ + cos²θ = 1
Double anglesin 2A = 2 sin A cos A; cos 2A = 1 − 2sin²A
Triple anglesin 3A = 3sin A − 4sin³A
Range of a sin θ + b cos θ[−√(a² + b²), √(a² + b²)]
General solution of sin θ = sin αnπ + (−1)ⁿα
Complementary inversessin−1x + cos−1x = π/2
tan−1 sum (xy < 1)tan−1((x + y)/(1 − xy))
Sine rulea/sin A = 2R
Cosine rulea² = b² + c² − 2bc cos A
Circumradius, inradiusR = abc/4Δ, r = Δ/s

8. Mind Map

Trigonometry
Ratios & identities
  • ASTC signs
  • Compound, double, triple angles
  • Sum ↔ product
Equations
  • nπ + (−1)ⁿα
  • 2nπ ± α
  • Counting solutions
Inverse functions
  • Principal ranges
  • sin−1 + cos−1 = π/2
  • tan−1 addition
Triangles
  • Sine, cosine rules
  • Area, Heron
  • R and r

9. Tips and Tricks

Shortcuts
  • For cos 20° cos 40° cos 80°, use cos θ cos(60° − θ) cos(60° + θ) = (1/4)cos 3θ with θ = 20°: the product = (1/4)cos 60° = 1/8.
  • For maximum/minimum of expressions like 3 sin x + 4 cos x + c, add ±√(a² + b²) to c.
  • To count solutions, sketch the graphs on the interval — intersections are the solutions.
  • For sin−1(sin x) and similar, draw the "zig-zag" graph or reduce x to the principal range using symmetry.
Mistakes students make
  • Dividing both sides of an equation by sin x or cos x and losing solutions where it is zero.
  • Writing sin−1(sin 2π/3) = 2π/3 — the correct value is π/3.
  • Squaring an equation and not rejecting the extra (spurious) solutions.

10. Important One-liners

  • sin θ and cos θ always lie between −1 and 1; tan θ can take any real value.
  • The period of sin x and cos x is 2π; of tan x, π; of |sin x|, π.
  • In any triangle, the largest side is opposite the largest angle.
  • sin 3θ = 3 sin θ − 4 sin³θ is the key to finding sin 18°.
  • In a right-angled triangle, the circumradius is half the hypotenuse.
  • tan A + tan B + tan C = tan A tan B tan C when A + B + C = π.

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Numerical

Find the number of solutions of sin 2x = cos x in [0, 2π].

Show solution

sin 2x − cos x = cos x (2 sin x − 1) = 0.

cos x = 0: x = π/2, 3π/2. sin x = 1/2: x = π/6, 5π/6.

These are 4 distinct solutions.

Answer: 4

Q2Easy

The maximum value of 3 sin x + 4 cos x is:

  1. 7
  2. 5
  3. 4
  4. 1
Show solution

Maximum = √(3² + 4²) = 5.

Answer: (B)

Q3Medium

tan−1(1/2) + tan−1(1/3) equals:

  1. π/6
  2. π/4
  3. π/3
  4. π/2
Show solution

Since (1/2)(1/3) < 1: sum = tan−1((1/2 + 1/3)/(1 − 1/6)) = tan−1((5/6)/(5/6)) = tan−11 = π/4.

Answer: (B)

Q4Numerical

Find the value of cos 20° cos 40° cos 80°.

Show solution

Using cos θ cos(60° − θ) cos(60° + θ) = (1/4)cos 3θ with θ = 20°: cos 20° cos 40° cos 80° = (1/4) cos 60° = 1/8 = 0.125.

Answer: 0.125

Q5Medium

The value of sin−1(sin 2π/3) is:

  1. 2π/3
  2. π/3
  3. −π/3
  4. π/6
Show solution

2π/3 is outside [−π/2, π/2]. Since sin(2π/3) = sin(π − 2π/3) = sin(π/3) and π/3 is in the principal range, the value is π/3.

Answer: (B)

Q6Easy

The general solution of sin x = 1/2 is:

  1. x = nπ + (−1)ⁿ π/6
  2. x = 2nπ ± π/6
  3. x = nπ + π/6
  4. x = nπ ± π/3
Show solution

sin x = sin(π/6), so x = nπ + (−1)ⁿ π/6.

Answer: (A)

Q7Medium

In a triangle with sides 3, 4 and 5, the circumradius is:

  1. 5/2
  2. 2
  3. 3
  4. 5
Show solution

3² + 4² = 5², so the triangle is right-angled and the hypotenuse is a diameter of the circumcircle: R = 5/2.

(Formula check: Δ = 6, R = abc/4Δ = 60/24 = 5/2.)

Answer: (A)

Q8Hard

If sin−1x + sin−1y = 2π/3, then cos−1x + cos−1y equals:

  1. π/3
  2. 2π/3
  3. π/6
  4. π
Show solution

cos−1x = π/2 − sin−1x and cos−1y = π/2 − sin−1y.

Sum = π − (sin−1x + sin−1y) = π − 2π/3 = π/3.

Answer: (A)

12. Quick Revision Summary

Remember before the exam
  • ASTC for signs; learn the 0°–90° table and sin 15°, sin 18°.
  • a sin θ + b cos θ ∈ [−√(a² + b²), √(a² + b²)].
  • sin θ = sin α ⇒ nπ + (−1)ⁿα; cos ⇒ 2nπ ± α; tan ⇒ nπ + α.
  • Principal ranges: sin−1 [−π/2, π/2], cos−1 [0, π], tan−1 (−π/2, π/2).
  • Sine rule = 2R; cosine rule; Δ = ½bc sin A; R = abc/4Δ, r = Δ/s.
All Mathematics chaptersJEE Main Mock Tests