Trigonometri AS/A level puremath 1A

Comprehensive Trigonometry Guide for A/AS Level Pure Mathematics

Trigonometry, a crucial branch of mathematics, explores the relationships between triangle sides and angles. This guide covers essential concepts, formulas, and applications for A/AS level students.

1. Fundamental Trigonometric Ratios

In a right-angled triangle with angle θ, the primary trigonometric ratios are defined as:

• Sine: sin θ = opposite / hypotenuse = y / r
• Cosine: cos θ = adjacent / hypotenuse = x / r
• Tangent: tan θ = opposite / adjacent = y / x

2. Critical Trigonometric Identities

Key identities include:

• Tangent identity: tan θ = sin θ / cos θ (when cos θ ≠ 0)
• Pythagorean identity: sin²θ + cos²θ = 1

3. Precise Trigonometric Function Values

Angle θ	sin θ	cos θ	tan θ
30° = π/6	1/2	√3/2	1/√3
45° = π/4	1/√2	1/√2	1
60° = π/3	√3/2	1/2	√3

4. Understanding Positive and Negative Angles

Positive angles are measured counterclockwise from the positive x-axis, while negative angles are measured clockwise. The mnemonic "All Students Take Calculus" helps recall where each function is positive in different quadrants.

5. Trigonometric Function Graphs

Understanding sine, cosine, and tangent graphs is crucial for solving equations and inequalities.

6. Trigonometric Function Transformations

Parameters affect trigonometric graphs as follows:

• y = a sin x: Vertical stretch by factor a
• y = sin (ax): Horizontal compression by factor 1/a
• y = a + sin x: Vertical translation by a units
• y = sin (x + a): Horizontal translation by -a units

7. Inverse Trigonometric Functions

Inverse functions are essential for equation solving:

• y = sin⁻¹ x: Domain: [-1, 1], Range: [-π/2, π/2]
• y = cos⁻¹ x: Domain: [-1, 1], Range: [0, π]
• y = tan⁻¹ x: Domain: All real numbers, Range: (-π/2, π/2)

8. Illustrative Problems

Problem 1: Solve sin x = 1/2 for 0 ≤ x < 2π

Solution:

1. sin 30° = 1/2, so x = 30° = π/6 is one solution
2. The other solution is 180° - 30° = 150° = 5π/6
3. Therefore, solutions are x = π/6 and x = 5π/6

Problem 2: Find the exact value of tan (60° + 45°)

Solution:

1. Use the tangent addition formula: tan (A + B) = (tan A + tan B) / (1 - tan A tan B)
2. tan 60° = √3 and tan 45° = 1
3. Substituting: tan (60° + 45°) = (√3 + 1) / (1 - √3 · 1) = (√3 + 1) / (1 - √3)
4. Rationalizing: ((√3 + 1)(1 + √3)) / ((1 - √3)(1 + √3)) = (4 + 2√3) / (-2) = -2 - √3

Conclusion

Mastering these trigonometric concepts is crucial for A/AS level mathematics success. Regular practice with diverse problems will enhance understanding and problem-solving skills. Explore additional resources to further develop your trigonometry knowledge.

Additional Example Problems

Problem 3: Solving a Trigonometric Equation

Question: Solve the equation 2 sin² x - sin x - 1 = 0 for 0 ≤ x < 2π.

Solution:

1. Let y = sin x
2. The equation becomes: 2y² - y - 1 = 0
3. This is a quadratic equation in y
4. Using the quadratic formula: y = (-b ± √(b² - 4ac)) / (2a)
5. y = (1 ± √(1 + 8)) / 4 = (1 ± 3) / 4
6. y = 1 or y = -1/2
7. So, sin x = 1 or sin x = -1/2
8. For sin x = 1, x = π/2
9. For sin x = -1/2, x = 7π/6 or 11π/6

Answer: The solutions are x = π/2, 7π/6, and 11π/6.

Problem 4: Trigonometric Identities

Question: Prove that (1 + tan² x) / (1 + cot² x) = sin² x.

Solution:

1. Start with the left side: (1 + tan² x) / (1 + cot² x)
2. Recall that tan x = sin x / cos x and cot x = cos x / sin x
3. Substitute: (1 + (sin² x / cos² x)) / (1 + (cos² x / sin² x))
4. Multiply numerator and denominator by sin² x cos² x: ((sin² x cos² x + sin⁴ x) / cos² x) / ((sin² x cos² x + cos⁴ x) / sin² x)
5. Simplify: (sin² x cos² x + sin⁴ x) / (sin² x cos² x + cos⁴ x)
6. Factor out sin² x from numerator and cos² x from denominator: sin² x (cos² x + sin² x) / cos² x (sin² x + cos² x)
7. Recall the Pythagorean identity: sin² x + cos² x = 1
8. Simplify: sin² x · 1 / cos² x · 1 = sin² x / cos² x
9. Multiply by cos² x / cos² x: sin² x

Answer: Thus, we have proven that (1 + tan² x) / (1 + cot² x) = sin² x.

Problem 5: Trigonometric Functions in Real-World Applications

Question: A Ferris wheel with a radius of 20 meters completes one revolution every 60 seconds. A rider boards at the bottom of the wheel. Write a function that describes the rider's height above the ground as a function of time t in seconds, and determine the rider's height after 15 seconds.

Solution:

1. The general form of the function will be: h(t) = a sin(bt) + c Where: a = amplitude (half the diameter) b = angular frequency (2π / period) c = vertical shift (distance from ground to center of wheel)
2. Amplitude: a = 20 meters
3. Angular frequency: b = 2π / 60 = π / 30 radians per second
4. Vertical shift: c = 20 meters (center of wheel)
5. The function is: h(t) = 20 sin((π/30)t) + 20
6. To find height at 15 seconds, substitute t = 15: h(15) = 20 sin((π/30)·15) + 20 = 20 sin(π/2) + 20 = 20 · 1 + 20 = 40 meters

Answer: The function describing the rider's height is h(t) = 20 sin((π/30)t) + 20, where h is in meters and t is in seconds. After 15 seconds, the rider's height is 40 meters above the ground.

Problem 6: Inverse Trigonometric Functions

Question: Evaluate the expression: sin(cos⁻¹(3/5)) + cos(sin⁻¹(4/5))

Solution:

1. Let's break this into two parts: A = sin(cos⁻¹(3/5)) and B = cos(sin⁻¹(4/5))
2. For part A: - In a right triangle where cos θ = 3/5, we have adjacent = 3 and hypotenuse = 5 - Using Pythagorean theorem, opposite = √(5² - 3²) = 4 - So, sin(cos⁻¹(3/5)) = 4/5
3. For part B: - In a right triangle where sin θ = 4/5, we have opposite = 4 and hypotenuse = 5 - Using Pythagorean theorem, adjacent = √(5² - 4²) = 3 - So, cos(sin⁻¹(4/5)) = 3/5
4. Adding the results: 4/5 + 3/5 = 7/5

Answer: The value of the expression is 7/5 or 1.4.

These additional problems cover a range of trigonometric concepts, including equation solving, identities, real-world applications, and inverse functions. They provide students with diverse practice opportunities to enhance their understanding and problem-solving skills in trigonometry.

Quiz 1

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Quiz 2:
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