Trigonometry Pure Math 2&3

Advanced Trigonometry Review

1. Basic Trigonometric Functions

Primary Functions:

• sine (sin θ)
• cosine (cos θ)
• tangent (tan θ)
• cosecant (cosec θ = 1/sin θ)
• secant (sec θ = 1/cos θ)
• cotangent (cot θ = 1/tan θ)

2. Important Trigonometric Identities

Pythagorean Identities:

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

Double Angle Formulas:

• sin 2θ = 2sinθ cosθ
• cos 2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ
• tan 2θ = 2tanθ/(1 - tan²θ)

3. Advanced Trigonometric Concepts

R-Method (R cos/sin(θ ± α)):

For expressions of the form asinθ + bcosθ:

• R = √(a² + b²)
• α = arctan(b/a)
• Result: R cos(θ - α) or R sin(θ + α)

4. Solving Trigonometric Equations

General Methods:

1. Express all terms using the same trigonometric function
2. Make substitutions to simplify complex expressions
3. Use factorization when possible
4. Remember to find all solutions in the given interval

Example: Solving 2tan²x + secx = 1

1. Replace secx with 1/cosx
2. Convert tan²x to (sin²x)/(cos²x)
3. Find common denominator
4. Solve for x in the specified interval

5. Special Angle Formulas

Values to Remember:

• sin 30° = 1/2
• cos 30° = √3/2
• sin 45° = 1/√2
• cos 45° = 1/√2
• sin 60° = √3/2
• cos 60° = 1/2

6. Compound Angle Formulas

• sin(A ± B) = sinA cosB ± cosA sinB
• cos(A ± B) = cosA cosB ∓ sinA sinB
• tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB)

7. Triple Angle Formulas

sin 3θ = 3sinθ - 4sin³θ

cos 3θ = 4cos³θ - 3cosθ

Trigonometry Practice Problems

Enhance Your Understanding of Trigonometry with These Practice Problems

A. Basic Trigonometric Functions

Problem 1 Easy

Solve the equation: 2sec(2x - 90°) = 3 for x ∈ [0°, 180°]

Solution:

Step 1: Isolate secant

sec(2x - 90°) = 3/2

Step 2: Take arccos of both sides

2x - 90° = ±arccos(2/3)

Step 3: Solve for x

2x = ±arccos(2/3) + 90°

x = (±arccos(2/3) + 90°)/2

Using calculator: x ≈ 48.19° or 131.81°

Answer: x = 48.19° or 131.81°

Problem 2 Medium

Express the equation cosec θ = 3sin θ + cot θ in terms of sin θ only, and solve for θ ∈ [0°, 180°]

Solution:

Step 1: Convert all terms to sin θ

• cosec θ = 1/sin θ

• cot θ = cos θ/sin θ = √(1-sin²θ)/sin θ

Step 2: Substitute

1/sin θ = 3sin θ + √(1-sin²θ)/sin θ

Step 3: Multiply all terms by sin θ

1 = 3sin²θ + √(1-sin²θ)

Step 4: Solve by substituting sin θ = x

1 = 3x² + √(1-x²)

x = ±0.5 (only positive value in first quadrant)

θ = 30° or 150°

Answer: θ = 30° or 150°

Problem 3 Hard

Prove that: sin(60° + x) + cos(30° + x) ≡ √3 cos x

Solution:

Step 1: Use compound angle formulas

sin(60° + x) = sin60°cosx + cos60°sinx

cos(30° + x) = cos30°cosx - sin30°sinx

Step 2: Substitute special angle values

sin60° = √3/2, cos60° = 1/2

sin30° = 1/2, cos30° = √3/2

Step 3: Expand

(√3/2)cosx + (1/2)sinx + (√3/2)cosx - (1/2)sinx

Step 4: Combine like terms

= √3cosx

Therefore, sin(60° + x) + cos(30° + x) ≡ √3 cos x

B. R-method and Advanced Problems

Problem 4 Medium

Express 3cos θ + 4sin θ in the form Rcos(θ - α), where R > 0 and α ∈ [0°, 90°]

Solution:

Step 1: Using R-method formula

R = √(3² + 4²) = √25 = 5

Step 2: Find α

tan α = 4/3

α = arctan(4/3) ≈ 53.13°

Step 3: Verify by expanding

5cos(θ - 53.13°) = 5(cosθcos53.13° + sinθsin53.13°)

= 3cosθ + 4sinθ

Answer: 5cos(θ - 53.13°)

Problem 5 Hard

Solve the equation: 2cot²x + cosecx = 1 for x ∈ [0°, 360°]

Solution:

Step 1: Let y = sinx

cosecx = 1/y

cot²x = cos²x/sin²x = (1-y²)/y²

Step 2: Substitute

2((1-y²)/y²) + 1/y = 1

Step 3: Multiply all terms by y²

2(1-y²) + y = y²

2 - 2y² + y = y²

2 = 3y² - y

3y² - y - 2 = 0

Step 4: Solve quadratic

y = (1 ± √(1 + 24))/6 = (1 ± 5)/6

y = 1 or -2/3

Step 5: Find x

x = arcsin(1) = 90°, 270°

x = arcsin(-2/3) ≈ 138.19°, 401.81°

Answer: x = 90°, 138.19°, 270°

Quiz 1

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