The Hook: The Circle That Rules JEE
Remember the Interstellar spinning spaceship scene? Or the Inception rotating hallway fight? Both involve circular motion, and guess what powers the math behind all rotations? Trigonometric identities!
Here’s the deal: In JEE, nearly every rotation problem, oscillation question, and complex number challenge traces back to these identities. Master them, and you’ve unlocked 15-20% of your JEE Math paper.
Question: Why do sin²θ + cos²θ always equal 1, no matter what angle you pick? By the end of this, you’ll not just know it—you’ll see it.
The Core Concept
What Are Trigonometric Identities?
In simple terms: Trigonometric identities are equations involving trig functions that are always true for all values where they’re defined. Think of them as the “multiplication tables” of trigonometry—once you know them, complex problems become simple.
The Fundamental Identity: Pythagoras in Disguise
$$\boxed{\sin^2\theta + \cos^2\theta = 1}$$Why it works: Drop a perpendicular from any point on the unit circle. By Pythagoras theorem: (opposite)² + (adjacent)² = (hypotenuse)². Since hypotenuse = 1 for unit circle, we get sin²θ + cos²θ = 1.
Visual proof:
Memory Tricks & Patterns
The “All Students Take Calculus” Rule
Memory Trick for Sign Convention:
“All Students Take Calculus” or “Add Sugar To Coffee”
- Quadrant I (0° to 90°): All functions positive
- Quadrant II (90° to 180°): Sine (and cosec) positive only
- Quadrant III (180° to 270°): Tan (and cot) positive only
- Quadrant IV (270° to 360°): Cos (and sec) positive only
The Three Sacred Identities
$$\boxed{ \begin{align} \sin^2\theta + \cos^2\theta &= 1 \\ 1 + \tan^2\theta &= \sec^2\theta \\ 1 + \cot^2\theta &= \csc^2\theta \end{align} }$$Mnemonic: “Some People Can”
- Sin² + Power 1 = Cos² (rearranged: sin² + cos² = 1)
- 1 + Tan² = Sec² (divide first identity by cos²θ)
- 1 + Cot² = Cosec² (divide first identity by sin²θ)
Compound Angle Formulas
Addition Formulas (The Building Blocks)
$$\boxed{ \begin{align} \sin(A + B) &= \sin A \cos B + \cos A \sin B \\ \sin(A - B) &= \sin A \cos B - \cos A \sin B \\ \cos(A + B) &= \cos A \cos B - \sin A \sin B \\ \cos(A - B) &= \cos A \cos B + \sin A \sin B \end{align} }$$Memory Trick: “Sine Changes, Cosine Doesn’t”
- For sin(A + B): both terms are additions (sin·cos + cos·sin)
- For cos(A + B): sign changes to minus (cos·cos - sin·sin)
Alternative Mnemonic: “SCP CPS” for sin(A+B) = SinCos + Plus CosSin
JEE Pattern: 60% of compound angle questions test sin(A+B) and cos(A-B) together!
Tangent Addition Formulas
$$\boxed{ \begin{align} \tan(A + B) &= \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ \tan(A - B) &= \frac{\tan A - \tan B}{1 + \tan A \tan B} \end{align} }$$Memory Trick: Notice the sign flip in denominator:
- Addition: minus in bottom
- Subtraction: plus in bottom
“Opposite Day in Denominator!”
Double Angle Formulas
The Power Trio
$$\boxed{ \begin{align} \sin 2\theta &= 2\sin\theta\cos\theta \\ \cos 2\theta &= \cos^2\theta - \sin^2\theta \\ &= 2\cos^2\theta - 1 \\ &= 1 - 2\sin^2\theta \\ \tan 2\theta &= \frac{2\tan\theta}{1-\tan^2\theta} \end{align} }$$Memory Trick for sin 2θ: “2 Sinners Cosplaying” → 2 sin cos
Memory Trick for cos 2θ three forms:
- C² - S² (difference of squares, basic form)
- 2C² - 1 (substitute S² = 1 - C²)
- 1 - 2S² (substitute C² = 1 - S²)
JEE Shortcut: If question has sin θ given, use form 3. If cos θ given, use form 2!
Triple Angle Formulas
$$\boxed{ \begin{align} \sin 3\theta &= 3\sin\theta - 4\sin^3\theta \\ \cos 3\theta &= 4\cos^3\theta - 3\cos\theta \\ \tan 3\theta &= \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \end{align} }$$Pattern Recognition: Notice the 3-4 pattern:
- sin 3θ: coefficient 3 for linear term, 4 for cubic
- cos 3θ: coefficient 4 for cubic, 3 for linear
- They’re flipped!
Memory Trick: “Sine starts with 3, Cosine starts with 4” (in decreasing power order)
Transformation Formulas (Product to Sum)
Product → Sum Conversions
$$\boxed{ \begin{align} \sin A \cos B &= \frac{1}{2}[\sin(A+B) + \sin(A-B)] \\ \cos A \sin B &= \frac{1}{2}[\sin(A+B) - \sin(A-B)] \\ \cos A \cos B &= \frac{1}{2}[\cos(A+B) + \cos(A-B)] \\ \sin A \sin B &= \frac{1}{2}[\cos(A-B) - \cos(A+B)] \end{align} }$$Memory Trick: “SC adds, CC adds, SS reverses”
- Sin·Cos: both sines with +
- Cos·Cos: both cosines with +
- Sin·Sin: both cosines but reversed signs (A-B first, then -)
Sum → Product Conversions
$$\boxed{ \begin{align} \sin C + \sin D &= 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right) \\ \sin C - \sin D &= 2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) \\ \cos C + \cos D &= 2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right) \\ \cos C - \cos D &= -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) \end{align} }$$Master Mnemonic: “Sums Start with Sin, Differences are Different”
Substitute C = A+B, D = A-B to derive product formulas!
JEE Time-Saver: These formulas convert products (hard to integrate/differentiate) into sums (easy)!
Half Angle Formulas
$$\boxed{ \begin{align} \sin\frac{\theta}{2} &= \pm\sqrt{\frac{1-\cos\theta}{2}} \\ \cos\frac{\theta}{2} &= \pm\sqrt{\frac{1+\cos\theta}{2}} \\ \tan\frac{\theta}{2} &= \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta} \end{align} }$$Memory Pattern:
- sin θ/2: has minus (1 - cos θ)
- cos θ/2: has plus (1 + cos θ)
JEE Trick: tan θ/2 has two forms—use whichever is given in the question!
When to Use Each Identity
Given sin θ or cos θ → Find other functions? → Use sin²θ + cos²θ = 1
Given tan θ → Find sec θ or vice versa? → Use 1 + tan²θ = sec²θ
Expression has 2θ → Simplify to θ? → Use double angle formulas
Need to find sin 15° or cos 75°? → Use sin(45° - 30°) or cos(45° + 30°) with compound angles
Product of two trig functions → Need to integrate/differentiate? → Convert product to sum first
Sum of two trig functions → Need to simplify? → Convert sum to product
Integration of form ∫ sin² x dx or ∫ cos² x dx? → Use cos 2x formulas to reduce power
Common Mistakes to Avoid
Wrong: cos(A + B) = cos A cos B + sin A sin B ❌
Right: cos(A + B) = cos A cos B - sin A sin B ✓
Why students mess up: They confuse it with sin(A + B) which has +.
Fix: Remember “Cosine Changes sign” → minus for addition.
Wrong: sin(θ/2) = √[(1 - cos θ)/2] without considering sign
Right: sin(θ/2) = ±√[(1 - cos θ)/2]
Fix: Check which quadrant θ/2 lies in using ASTC rule!
Wrong: tan 2θ = 2 tan θ / (1 + tan²θ)
Right: tan 2θ = 2 tan θ / (1 - tan²θ)
Why it matters: Gets you to undefined at θ = 45° (which is correct behavior).
Common Error: Using wrong formula for sin A sin B
Wrong: sin A sin B = ½[cos(A-B) + cos(A+B)]
Right: sin A sin B = ½[cos(A-B) - cos(A+B)]
Mnemonic: “SS is the rebel—it has reversed sign!”
Practice Problems
Level 1: Foundation (NCERT)
Question: If sin θ = 3/5 and θ is in the second quadrant, find cos θ and tan θ.
Solution: Step 1: Use sin²θ + cos²θ = 1
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 16/25
cos θ = ±4/5
Step 2: Since θ is in quadrant II, cos θ is negative (by ASTC rule)
cos θ = -4/5
Step 3: tan θ = sin θ / cos θ
tan θ = (3/5) / (-4/5) = -3/4
Answer: cos θ = -4/5, tan θ = -3/4
Question: Find the exact value of sin 75°.
Solution: Step 1: Write 75° = 45° + 30° Step 2: Apply sin(A + B) = sin A cos B + cos A sin B
sin 75° = sin(45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
= (√2/2)(√3/2) + (√2/2)(1/2)
= (√6/4) + (√2/4)
= (√6 + √2)/4
Answer: sin 75° = (√6 + √2)/4
Question: If tan θ = 2, find tan 2θ.
Solution: Use tan 2θ = 2 tan θ / (1 - tan²θ)
tan 2θ = 2(2) / (1 - 2²)
= 4 / (1 - 4)
= 4 / (-3)
= -4/3
Answer: tan 2θ = -4/3
Level 2: JEE Main
Question: Prove that: (sin 3θ + sin θ)/(cos 3θ + cos θ) = tan 2θ
Solution: Use sum-to-product formulas:
Numerator:
sin 3θ + sin θ = 2 sin[(3θ+θ)/2] cos[(3θ-θ)/2]
= 2 sin 2θ cos θ
Denominator:
cos 3θ + cos θ = 2 cos[(3θ+θ)/2] cos[(3θ-θ)/2]
= 2 cos 2θ cos θ
Therefore:
(sin 3θ + sin θ)/(cos 3θ + cos θ) = (2 sin 2θ cos θ)/(2 cos 2θ cos θ)
= sin 2θ / cos 2θ
= tan 2θ
Proved!
Question: If sin θ + cos θ = √2, find sin⁴θ + cos⁴θ.
Solution: Step 1: Square both sides
(sin θ + cos θ)² = 2
sin²θ + cos²θ + 2 sin θ cos θ = 2
1 + 2 sin θ cos θ = 2
sin θ cos θ = 1/2
Step 2: Find sin⁴θ + cos⁴θ using identity
sin⁴θ + cos⁴θ = (sin²θ + cos²θ)² - 2 sin²θ cos²θ
= 1² - 2(sin θ cos θ)²
= 1 - 2(1/2)²
= 1 - 2(1/4)
= 1 - 1/2
= 1/2
Answer: 1/2
Question: Express 2 sin 5x cos 3x as a sum.
Solution: Use sin A cos B = ½[sin(A+B) + sin(A-B)]
Here A = 5x, B = 3x:
2 sin 5x cos 3x = 2 × ½[sin(5x+3x) + sin(5x-3x)]
= sin 8x + sin 2x
Answer: sin 8x + sin 2x
JEE Trick: Factor of 2 cancels with ½ from formula!
Level 3: JEE Advanced
Question: If α + β = π/4, prove that: (1 + tan α)(1 + tan β) = 2
Solution: Expand LHS:
(1 + tan α)(1 + tan β) = 1 + tan α + tan β + tan α tan β
Since α + β = π/4:
tan(α + β) = tan(π/4) = 1
Using tan(α + β) = (tan α + tan β)/(1 - tan α tan β):
1 = (tan α + tan β)/(1 - tan α tan β)
1 - tan α tan β = tan α + tan β
tan α + tan β = 1 - tan α tan β
Substitute into LHS:
1 + (tan α + tan β) + tan α tan β = 1 + (1 - tan α tan β) + tan α tan β
= 1 + 1
= 2
Proved!
Question: If cos(α - β) = 1 and cos(α + β) = 1/e, where α, β ∈ [-π, π], find the number of ordered pairs (α, β).
Solution: From cos(α - β) = 1:
α - β = 0 (since cos 0 = 1 and α - β ∈ [-2π, 2π])
α = β
From cos(α + β) = 1/e:
α + β = ± cos⁻¹(1/e)
Since α = β:
2α = ± cos⁻¹(1/e)
α = ± cos⁻¹(1/e)/2
Since α ∈ [-π, π] and 1/e ≈ 0.368:
cos⁻¹(1/e) ≈ 1.19 radians
Both α = cos⁻¹(1/e)/2 and α = -cos⁻¹(1/e)/2 lie in [-π, π].
Since β = α, we have 2 ordered pairs:
- (cos⁻¹(1/e)/2, cos⁻¹(1/e)/2)
- (-cos⁻¹(1/e)/2, -cos⁻¹(1/e)/2)
Answer: 2 ordered pairs
Question: Find the maximum value of sin x + cos x.
Solution: Method 1 (Using compound angles):
sin x + cos x = √2 [sin x/√2 + cos x/√2]
= √2 [sin x cos 45° + cos x sin 45°]
= √2 sin(x + 45°)
Maximum value of sin(x + 45°) = 1
Therefore, maximum of sin x + cos x = √2(1) = √2
Method 2 (Calculus - bonus):
Let y = sin x + cos x
dy/dx = cos x - sin x = 0
tan x = 1
x = 45°
y_max = sin 45° + cos 45° = √2/2 + √2/2 = √2
Answer: √2 (achieved at x = 45° + 360°n)
Quick Revision Box
| Identity Type | Formula | When to Use |
|---|---|---|
| Pythagorean | sin²θ + cos²θ = 1 | Finding one from other |
| Pythagorean | 1 + tan²θ = sec²θ | Tan-Sec conversions |
| Double Angle | sin 2θ = 2 sin θ cos θ | Simplifying 2θ to θ |
| Double Angle | cos 2θ = 1 - 2sin²θ | Integration of sin²θ |
| Compound | sin(A±B) = sin A cos B ± cos A sin B | Finding exact values |
| Compound | cos(A±B) = cos A cos B ∓ sin A sin B | Simplification |
| Product→Sum | sin A cos B = ½[sin(A+B) + sin(A-B)] | Integration of products |
| Sum→Product | sin C + sin D = 2sin((C+D)/2)cos((C-D)/2) | Solving equations |
| Half Angle | tan(θ/2) = sin θ/(1+cos θ) | Simplifying half angles |
Links to Related Topics
Connected Concepts within Trigonometry:
- Trigonometric Equations - Apply these identities to solve equations
- Inverse Trigonometric Functions - Domain/range uses these
- Properties of Triangles - Apply to triangle geometry
Applications in Calculus:
- Trigonometric Integration - Uses product-to-sum heavily
- Integration Techniques - Trig substitution
- Indefinite Integrals - Integrating trig functions
- Differentiation Rules - Derivatives of trig functions
Complex Numbers Connection:
- Complex Numbers Basics - e^(i theta) = cos theta + i sin theta
- De Moivre’s Theorem - Powers of complex numbers
- Argand Diagram - Polar form uses trig
Physics Applications:
- Simple Harmonic Motion - SHM involves trig identities
- Circular Motion - Angular motion uses trig
- Projectile Motion - Trajectory analysis
What’s Next:
- Trigonometric Equations - Now use these identities to solve equations!
Teacher’s Summary
The Three Pillars: Master sin²θ + cos²θ = 1 and its two derivative forms (divide by cos²θ and sin²θ respectively). These unlock 80% of identity problems.
Compound Angles = Building Blocks: Every other identity (double, triple, half-angle) can be derived from sin(A±B) and cos(A±B). Don’t memorize blindly—understand the derivation.
Transformation Formulas Save Time: Product-to-sum and sum-to-product are NOT for memorization—they’re for integration and equation-solving. Keep the formula sheet handy initially.
Sign Convention is Everything: Use ASTC (“All Students Take Calculus”) religiously. Most JEE wrong answers come from sign errors.
Pattern Over Memory: Notice 3-4 pattern in triple angles, sign flip in compound angles. JEE rewards pattern recognition over rote learning.
Exam Strategy: In a 3-hour paper, you’ll save 10-15 minutes if you can recall these identities instantly. That’s worth 2-3 extra questions attempted!
“Trigonometry is not about memorizing 30 formulas—it’s about mastering 5 core identities and knowing how to transform them. The unit circle never lies.”
Weightage:
- JEE Main: 2-3 questions directly, plus 5-6 questions indirectly (in calculus, complex numbers)
- JEE Advanced: 1-2 questions directly, heavily used in geometry and calculus
Time-Saving Tip: Create a single A4 formula sheet with only the “core 5” identities. Derive others on the fly—builds confidence and saves memory load!