Inequalities and Optimization

🎢 Trigonometric Inequalities & Optimization

The Roller Coaster Analogy 🎡

Imagine you’re designing a roller coaster. The tracks go up and down in smooth waves—just like sine and cosine functions! Sometimes you need to know:

• When is the coaster above a certain height? (Inequalities)
• What’s the highest point? (Maximum)
• What’s the lowest point? (Minimum)
• What heights can it reach? (Range)

This is exactly what we’ll learn today!

1. Trigonometric Inequalities 📊

What Are They?

Instead of solving sin(x) = 0.5, we solve things like sin(x) > 0.5 or cos(x) ≤ -0.3.

Think of it like this: Instead of asking “When is the roller coaster at exactly 50 meters?”, we ask “When is it ABOVE 50 meters?”

Key Insight 💡

Trig functions repeat! So solutions come in intervals that repeat every period.

Simple Example

Solve: sin(x) > 0 for x ∈ [0, 2π]

Think: When is sine positive?

• Sine is positive in Quadrants I and II
• That means: x ∈ (0, π)

Solution: x ∈ (0, π)

Another Example

Solve: cos(x) ≤ 0.5 for x ∈ [0, 2π]

Step 1: Find where cos(x) = 0.5

• x = π/3 and x = 5π/3

Step 2: Cosine ≤ 0.5 between these points

• x ∈ [π/3, 5π/3]

graph TD
    A["cos x = 0.5"] --> B["x = π/3"]
    A --> C["x = 5π/3"]
    B --> D["Check intervals"]
    C --> D
    D --> E["x ∈ π/3, 5π/3"]

2. Wavy Curve Method 🌊

The Magic Trick

This is your superpower for solving polynomial-type trig inequalities!

How It Works

Step 1: Find all zeros (roots)

Step 2: Mark them on a number line

Step 3: Start from the RIGHT with a positive sign

Step 4: Alternate signs at each zero

Example

Solve: (sin x - 0.5)(cos x + 1) > 0 for x ∈ [0, 2π]

Step 1: Find zeros

• sin x = 0.5 → x = π/6, 5π/6
• cos x = -1 → x = π

Step 2: Mark on number line: π/6, 5π/6, π

Step 3: Draw the wavy curve

      +        -        +        -
  ----●--------●--------●--------●----
      π/6     5π/6      π       2π

Solution: x ∈ (0, π/6) ∪ (5π/6, π) ∪ (π, 2π)

Why It Works 🧠

Each factor changes sign when it equals zero. Like a light switch flipping on and off!

3. Maximum of Trig Expressions 📈

The Golden Rules

Example 1: Simple

Find max of: 3 sin x

Answer: 3 × 1 = 3 (when sin x = 1)

Example 2: Combined Expression

Find max of: 3 sin x + 4 cos x

Using the formula: √(3² + 4²) = √25 = 5

The Secret 🔮

Any expression a·sin x + b·cos x has maximum √(a² + b²)

This works because we can rewrite it as a single trig function!

4. Minimum of Trig Expressions 📉

The Mirror Image

Example

Find min of: 5 sin x - 12 cos x

Answer: -√(5² + 12²) = -√169 = -13

Pro Tip 💪

The minimum is just the negative of the maximum for expressions of type a·sin x + b·cos x.

5. Range of Trig Expressions 🎯

What Is Range?

The range is ALL possible values a function can output.

Basic Ranges

Example: Finding Range

Find range of: f(x) = 2 sin x + 3

Step 1: Range of sin x is [-1, 1]

Step 2: Multiply by 2: [-2, 2]

Step 3: Add 3: [1, 5]

Range: [1, 5]

graph TD
    A["sin x range: -1 to 1"] --> B["×2: -2 to 2"]
    B --> C["+3: 1 to 5"]
    C --> D["Final Range: 1, 5"]

Another Example

Find range of: f(x) = sin x · cos x

Use identity: sin x · cos x = (1/2) sin 2x

Range of sin 2x: [-1, 1]

Multiply by 1/2: [-1/2, 1/2]

Range: [-1/2, 1/2]

6. Auxiliary Angle Method 🎭

The Name-Changer

This method transforms a·sin x + b·cos x into a single trig function!

The Formula

a·sin x + b·cos x = R·sin(x + φ)

Where:

• R = √(a² + b²)
• tan φ = b/a

Example

Convert: 3 sin x + 4 cos x to R·sin(x + φ)

Step 1: Find R

• R = √(3² + 4²) = √25 = 5

Step 2: Find φ

• tan φ = 4/3
• φ = arctan(4/3) ≈ 53.13°

Result: 5 sin(x + 53.13°)

Why Is This Useful? 🤔

Now we can easily find:

• Maximum: 5 (when sin(x + φ) = 1)
• Minimum: -5 (when sin(x + φ) = -1)
• Range: [-5, 5]
• Zeros: When sin(x + φ) = 0

Alternative Form

You can also write:

a·sin x + b·cos x = R·cos(x - θ)

Where tan θ = a/b

7. Conditional Identities 🔗

What Are They?

These are special identities that work only when certain conditions are met.

The Most Famous Condition

If A + B + C = π (like angles in a triangle):

Example

If A + B + C = π, prove: tan A + tan B + tan C = tan A · tan B · tan C

Proof:

Since A + B + C = π → A + B = π - C → tan(A + B) = tan(π - C) = -tan C

Using addition formula:

(tan A + tan B)/(1 - tan A · tan B) = -tan C

Rearranging:

tan A + tan B = -tan C(1 - tan A · tan B)
tan A + tan B = -tan C + tan A · tan B · tan C
tan A + tan B + tan C = tan A · tan B · tan C ✓

Another Useful Identity

If A + B = 45°:

• (1 + tan A)(1 + tan B) = 2

Example

If α + β = π/4, find: (1 + tan α)(1 + tan β)

Solution: Since α + β = π/4 → tan(α + β) = 1

Using formula:

(tan α + tan β)/(1 - tan α · tan β) = 1
tan α + tan β = 1 - tan α · tan β

Now expand (1 + tan α)(1 + tan β):

= 1 + tan α + tan β + tan α · tan β
= 1 + (1 - tan α · tan β) + tan α · tan β
= 1 + 1
= 2 ✓

🎯 Quick Summary

🚀 You’ve Got This!

Remember our roller coaster? Now you can:

• ✅ Find when it’s above any height (inequalities)
• ✅ Find the peak (maximum)
• ✅ Find the valley (minimum)
• ✅ Know all possible heights (range)
• ✅ Simplify complex wave patterns (auxiliary angle)
• ✅ Use special shortcuts for triangles (conditional identities)

The wavy curve is your friend. The auxiliary angle is your superpower. Go conquer those trig problems! 🎢