IB Mathematics AI – Topic 3

Unit Circle Equation:

\[ x^2 + y^2 = 1 \]

Trigonometric Ratios on Unit Circle:

For angle θ measured from positive x-axis, point P on unit circle has coordinates:

\[ P = (\cos\theta, \sin\theta) \]

x-coordinate: \(\cos\theta\)

y-coordinate: \(\sin\theta\)

Gradient (slope): \(\tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta}\)

Key Angle Values (memorize these!):

• 0° (0 rad): \(\cos 0 = 1\), \(\sin 0 = 0\)
• 30° (π/6): \(\cos 30° = \frac{\sqrt{3}}{2}\), \(\sin 30° = \frac{1}{2}\)
• 45° (π/4): \(\cos 45° = \frac{\sqrt{2}}{2}\), \(\sin 45° = \frac{\sqrt{2}}{2}\)
• 60° (π/3): \(\cos 60° = \frac{1}{2}\), \(\sin 60° = \frac{\sqrt{3}}{2}\)
• 90° (π/2): \(\cos 90° = 0\), \(\sin 90° = 1\)

Four Quadrants - CAST Rule:

• Quadrant I (0° to 90°): All positive - All
• Quadrant II (90° to 180°): Sin positive - Students
• Quadrant III (180° to 270°): Tan positive - Take
• Quadrant IV (270° to 360°): Cos positive - Calculus

⚠️ Common Pitfalls & Tips:

• Angles measured anticlockwise from positive x-axis
• Remember CAST: All-Students-Take-Calculus (counterclockwise from Quadrant I)
• cos corresponds to x, sin corresponds to y (not the other way!)
• Use reference angles in first quadrant, then adjust sign based on quadrant

1. Pythagorean Identity (most important!):

\[ \sin^2\theta + \cos^2\theta = 1 \]

Can be rearranged to:

\[ \sin^2\theta = 1 - \cos^2\theta \]

\[ \cos^2\theta = 1 - \sin^2\theta \]

2. Quotient Identity:

\[ \tan\theta = \frac{\sin\theta}{\cos\theta} \]

3. Reciprocal Identities:

\[ \csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta} \]

4. Complementary Angle Identity:

\[ \sin\theta = \cos(90° - \theta) \]

\[ \cos\theta = \sin(90° - \theta) \]

5. Double Angle Formulas (in IB formula booklet):

\[ \sin(2\theta) = 2\sin\theta\cos\theta \]

\[ \cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta \]

⚠️ Common Pitfalls & Tips:

• Pythagorean identity: \(\sin^2\theta + \cos^2\theta = 1\) NOT \(\sin\theta + \cos\theta = 1\)
• \(\sin^2\theta\) means \((\sin\theta)^2\), not \(\sin(\theta^2)\)
• Use identities to simplify before solving equations
• All identities in formula booklet - don't memorize unnecessarily

Why is it Ambiguous?

When solving for an angle using sine rule, \(\sin\theta\) can have two values in range 0° to 180°:

• One in Quadrant I (acute angle)
• One in Quadrant II (obtuse angle)

Example:

If \(\sin\theta = 0.5\), then \(\theta = 30°\) or \(\theta = 150°\)

Finding Both Solutions:

Step 1: Use sine rule to find first angle (calculator gives acute angle)

\[ \theta_1 = \sin^{-1}(value) \]

Step 2: Find second possible angle using symmetry:

\[ \theta_2 = 180° - \theta_1 \]

Step 3: Check if both angles are valid (angles in triangle must sum to 180°)

Possible Outcomes:

• No solution: If calculated sine value > 1
• One solution: If one angle makes triangle impossible (sum > 180°)
• Two solutions: If both angles create valid triangles

⚠️ Common Pitfalls & Tips:

• Always check for second solution when using sine rule with SSA
• Calculator only gives acute angle - must find obtuse angle yourself
• Verify both solutions by checking angles sum to 180°
• Draw diagrams to visualize both possible triangles

Solution:

Given: a = 8, b = 10, A = 40° (SSA case - ambiguous!)

Step 1: Use sine rule

\[ \frac{\sin A}{a} = \frac{\sin B}{b} \]

\[ \frac{\sin 40°}{8} = \frac{\sin B}{10} \]

\[ \sin B = \frac{10 \times \sin 40°}{8} \approx \frac{10 \times 0.6428}{8} \approx 0.8035 \]

Step 2: Find first solution (acute)

\[ B_1 = \sin^{-1}(0.8035) \approx 53.5° \]

Step 3: Find second solution (obtuse)

\[ B_2 = 180° - 53.5° = 126.5° \]

Step 4: Check validity

For B₁ = 53.5°:

A + B₁ = 40° + 53.5° = 93.5° < 180° ✓ Valid

For B₂ = 126.5°:

A + B₂ = 40° + 126.5° = 166.5° < 180° ✓ Valid

Answer: B = 53.5° or B = 126.5° (both solutions valid)

General Steps for Solving Trig Equations:

1. Rearrange: Isolate the trig function (e.g., get \(\sin x = k\))
2. Find principal value: Use inverse function on calculator
3. Find other solutions: Use unit circle and symmetry
4. Apply periodicity: Add/subtract multiples of period
5. Check interval: Only include solutions in given range

For sin x = k:

Principal value: \(x_1 = \sin^{-1}(k)\)

Second value in [0°, 360°]: \(x_2 = 180° - x_1\)

General solution: \(x = x_1 + 360n°\) or \(x = x_2 + 360n°\), where n is integer

For cos x = k:

Principal value: \(x_1 = \cos^{-1}(k)\)

Second value in [0°, 360°]: \(x_2 = 360° - x_1\)

For tan x = k:

Principal value: \(x_1 = \tan^{-1}(k)\)

Second value in [0°, 360°]: \(x_2 = x_1 + 180°\)

(tan has period of 180°)

⚠️ Common Pitfalls & Tips:

• Don't stop at calculator answer - find ALL solutions in interval
• Use unit circle to visualize solutions
• Check calculator mode (degrees or radians)
• For equations with \(\sin^2x\) or \(\cos^2x\), use identities first

Solution:

Step 1: Isolate sin x

\[ \sin x = \frac{1}{2} \]

Step 2: Find principal value

\[ x_1 = \sin^{-1}(0.5) = 30° \]

Step 3: Find second solution using symmetry

For sine, second solution in [0°, 360°]:

\[ x_2 = 180° - 30° = 150° \]

Step 4: Check for additional solutions

Adding 360°: 30° + 360° = 390° (outside range)

Adding 360°: 150° + 360° = 510° (outside range)

Answer: x = 30° or x = 150°

Sine Function: y = sin x

• Domain: All real numbers \((-\infty, \infty)\)
• Range: \([-1, 1]\)
• Period: 360° (or \(2\pi\) radians)
• Amplitude: 1
• Starts at: (0, 0)
• Zeros: 0°, 180°, 360°, ...
• Maximum: 1 at 90°, 450°, ...
• Minimum: -1 at 270°, 630°, ...

Cosine Function: y = cos x

• Domain: All real numbers \((-\infty, \infty)\)
• Range: \([-1, 1]\)
• Period: 360° (or \(2\pi\) radians)
• Amplitude: 1
• Starts at: (0, 1)
• Zeros: 90°, 270°, 450°, ...
• Note: Cosine is sine shifted left by 90°

Tangent Function: y = tan x

• Domain: All real numbers except \(x = 90° + 180n°\)
• Range: All real numbers \((-\infty, \infty)\)
• Period: 180° (or \(\pi\) radians)
• No amplitude (unbounded)
• Vertical asymptotes: at 90°, 270°, ...
• Zeros: 0°, 180°, 360°, ...

General Form: y = A sin(B(x - C)) + D

• A: Amplitude (vertical stretch)
• B: Affects period: \(\text{Period} = \frac{360°}{B}\)
• C: Horizontal shift (phase shift)
• D: Vertical shift (moves midline)

⚠️ Common Pitfalls & Tips:

• Amplitude is |A|, not 2A (that's peak-to-peak distance)
• Period formula: \(\frac{360°}{B}\), not \(\frac{B}{360°}\)
• Tangent has period 180°, not 360°
• Use GDC to sketch and verify graphs

Unit Circle

• x = cos θ, y = sin θ
• CAST for signs
• Key angles: 30°, 45°, 60°, 90°

Key Identity

• \(\sin^2θ + \cos^2θ = 1\)
• \(\tan θ = \frac{\sin θ}{\cos θ}\)

Ambiguous Case

• SSA: Check both solutions
• θ₂ = 180° - θ₁
• Verify angles sum < 180°

Graphs

• Period: 360° (sin, cos)
• Period: 180° (tan)
• Amplitude = |A|