MYP Inquiry Framework
Key Concept Form & Relationships
Related Concepts Measurement, Generalization, Space
Global Context Scientific and technical innovation (Periodic motion, orbital mechanics & wave physics)
Statement of Inquiry Measuring circular displacement through natural radian ratios connects rotational geometry with continuous functional relationships.
4.1

Radian Measure & Degree-Radian Conversion

  • Definition of a Radian: One radian ($1\text{ rad}$) is the measure of the central angle subtended in a circle by an arc whose length equals the radius of the circle ($s = r$). $$\text{Full Revolution} = 360^\circ = 2\pi\text{ radians} \implies \mathbf{180^\circ = \pi\text{ radians}}$$ $$1\text{ rad} = \frac{180^\circ}{\pi} \approx 57.296^\circ, \qquad 1^\circ = \frac{\pi}{180}\text{ radians}$$
  • Key Benchmark Conversions: $$30^\circ = \frac{\pi}{6}, \quad 45^\circ = \frac{\pi}{4}, \quad 60^\circ = \frac{\pi}{3}, \quad 90^\circ = \frac{\pi}{2}, \quad 180^\circ = \pi, \quad 270^\circ = \frac{3\pi}{2}, \quad 360^\circ = 2\pi$$
Unit Circle: (cos θ, sin θ) P(x, y) = (cos θ, sin θ) θ The CAST Quadrant Rule A All positive S Sin positive T Tan positive C Cos positive
4.2

Arc Length, Sector Area & Segment Area

  • Arc Length ($s$): When the central angle $\theta$ is in radians: $$\mathbf{s = r\theta}$$
  • Area of a Circular Sector ($A$): $$\mathbf{A_{\text{sector}} = \frac{1}{2}r^2\theta}$$
  • Area of a Circular Segment: The region bounded by a chord and its corresponding arc: $$\mathbf{A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta)}$$ Critical reminder: On calculators, evaluate $\sin \theta$ with the angle $\theta$ in radian mode!
4.3

The Unit Circle & The CAST Quadrant Rule

  • Unit Circle Definition ($x^2 + y^2 = 1$): For any angle $\theta$ in standard position (vertex at origin, initial ray along positive $x$-axis): $$\mathbf{x = \cos \theta, \qquad y = \sin \theta, \qquad \tan \theta = \frac{y}{x} = \frac{\sin \theta}{\cos \theta}}$$
  • The CAST Rule (Signs Across Quadrants):
    • Quadrant 1 ($0 < \theta < \frac{\pi}{2}$): All functions ($\sin, \cos, \tan$) are positive.
    • Quadrant 2 ($\frac{\pi}{2} < \theta < \pi$): Only Sine is positive ($\cos, \tan$ are negative).
    • Quadrant 3 ($\pi < \theta < \frac{3\pi}{2}$): Only Tangent is positive ($\sin, \cos$ are negative).
    • Quadrant 4 ($\frac{3\pi}{2} < \theta < 2\pi$): Only Cosine is positive ($\sin, \tan$ are negative).
  • Reference Angles ($\theta_{\text{ref}}$): The acute angle formed between the terminal arm and the $x$-axis:
    • Q2: $\theta_{\text{ref}} = \pi - \theta$
    • Q3: $\theta_{\text{ref}} = \theta - \pi$
    • Q4: $\theta_{\text{ref}} = 2\pi - \theta$

Worked Examples (Criterion A & C)

Example 1 Sector and Segment Area Calculations

Problem: A circle with radius $r = 8\text{ cm}$ has a central sector subtended by an angle of $\theta = \frac{2\pi}{3}\text{ radians}$.
(a) Find the exact arc length.
(b) Find the area of the sector.
(c) Find the exact area of the circular segment.

Step 1: Calculate arc length $$s = r\theta = 8 \times \frac{2\pi}{3} = \mathbf{\frac{16\pi}{3}\text{ cm}} \approx 16.76\text{ cm}$$
Step 2: Calculate sector area $$A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(8^2)\left(\frac{2\pi}{3}\right) = \frac{1}{2}(64)\left(\frac{2\pi}{3}\right) = \mathbf{\frac{64\pi}{3}\text{ cm}^2} \approx 67.02\text{ cm}^2$$
Step 3: Calculate segment area $$A_{\triangle} = \frac{1}{2}r^2 \sin\theta = \frac{1}{2}(64)\sin\left(\frac{2\pi}{3}\right) = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3}\text{ cm}^2$$ $$A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} = \mathbf{\frac{64\pi}{3} - 16\sqrt{3}\text{ cm}^2} \approx 39.3\text{ cm}^2$$ Criterion A: Complete exact radical and $\pi$ formulation
Example 2 Unit Circle & CAST Exact Trigonometric Evaluation

Problem: Find the exact value of $\cos\left(\frac{5\pi}{4}\right)$ and $\tan\left(\frac{5\pi}{4}\right)$ using the unit circle.

Step 1: Identify quadrant and reference angle

Angle $\theta = \frac{5\pi}{4} = \pi + \frac{\pi}{4}$. This lies in Quadrant 3.

Reference angle: $\theta_{\text{ref}} = \frac{5\pi}{4} - \pi = \mathbf{\frac{\pi}{4}}$.

Step 2: Determine sign from CAST rule

In Quadrant 3 (T), only tangent is positive; cosine is negative.

Step 3: Evaluate exact values $$\cos\left(\frac{5\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = \mathbf{-\frac{\sqrt{2}}{2}}$$ $$\tan\left(\frac{5\pi}{4}\right) = +\tan\left(\frac{\pi}{4}\right) = \mathbf{1}$$ Criterion C: Explicit quadrant breakdown and reference angle deduction

MYP Extended Practice Exercises: Radians & Unit Circle

Tier 1 (Conversions & Arc Lengths), Tier 2 (Sectors, Segments & CAST Values), and Tier 3 (Extended Geometric Modeling).

Question 1 Tier 1: Standard 3 Marks

Convert the angles:
(a) $150^\circ$ into exact radian measure in terms of $\pi$.
(b) $\dfrac{7\pi}{6}\text{ radians}$ into degrees.

View Full Method Markscheme
[M1] Part (a): Multiply by $\frac{\pi}{180}$: $150 \times \frac{\pi}{180} = \mathbf{\frac{5\pi}{6}\text{ rad}}$.
[M1] Part (b): Multiply by $\frac{180}{\pi}$: $\frac{7\pi}{6} \times \frac{180}{\pi} = 7 \times 30$.
[A1] Calculate: $\mathbf{210^\circ}$.
Question 2 Tier 1: Standard 3 Marks

A circular arc has radius $r = 15\text{ cm}$ and subtends a central angle of $1.2\text{ radians}$. Find:
(a) The arc length.
(b) The area of the sector.

View Full Method Markscheme
[M1] Part (a): $s = r\theta = 15 \times 1.2 = \mathbf{18\text{ cm}}$.
[M1] Part (b): $A = \frac{1}{2}r^2\theta = \frac{1}{2}(15^2)(1.2) = \frac{1}{2}(225)(1.2)$.
[A1] Calculate: $A = 225 \times 0.6 = \mathbf{135\text{ cm}^2}$.
Question 3 Tier 1: Standard 3 Marks

Find the exact value of $\sin\left(\frac{5\pi}{6}\right)$ without using a calculator.

View Full Method Markscheme
[M1] $\frac{5\pi}{6}$ is in Quadrant 2. In Q2, sine is positive.
[M1] Reference angle: $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.
[A1] $\sin\left(\frac{5\pi}{6}\right) = +\sin\left(\frac{\pi}{6}\right) = \mathbf{\frac{1}{2}}$.
Question 4 Tier 2: Intermediate 4 Marks

A circle has a sector with perimeter $28\text{ cm}$ and area $48\text{ cm}^2$. Find the two possible values of the radius $r$.

View Full Method Markscheme
[M1] Sector perimeter consists of two radii plus the arc: $P = 2r + r\theta = 28 \implies r\theta = 28 - 2r$.
[M1] Sector area: $A = \frac{1}{2}r^2\theta = \frac{1}{2}r(r\theta) = 48$.
[M1] Substitute $r\theta$: $\frac{1}{2}r(28 - 2r) = 48 \implies r(14 - r) = 48 \implies 14r - r^2 = 48$.
[A1] Quadratic: $r^2 - 14r + 48 = 0 \implies (r - 6)(r - 8) = 0$. Possible radii: $\mathbf{r = 6\text{ cm} \quad \text{or} \quad r = 8\text{ cm}}$.
Question 5 Tier 2: Intermediate 4 Marks

If $\sin \theta = -\dfrac{3}{5}$ and $\theta$ lies in Quadrant 3, find the exact values of:
(a) $\cos \theta$
(b) $\tan \theta$

View Full Method Markscheme
[M1] In Quadrant 3 (T), cosine is negative and tangent is positive.
[M1] Reference triangle has opposite $= 3$, hypotenuse $= 5 \implies$ adjacent $= 4$.
[A1] Part (a): $\cos \theta = \mathbf{-\dfrac{4}{5}}$.
[A1] Part (b): $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-3/5}{-4/5} = \mathbf{+\dfrac{3}{4}}$.
Question 6 Tier 2: Intermediate 4 Marks

Find the area of the minor circular segment formed by a chord subtending an angle of $\theta = \frac{\pi}{3}\text{ radians}$ in a circle of radius $10\text{ cm}$. Give your answer to 3 significant figures.

View Full Method Markscheme
[M1] Segment formula: $A_{\text{seg}} = \frac{1}{2}r^2(\theta - \sin \theta)$.
[M1] Substitute $r = 10$ and $\theta = \frac{\pi}{3}$: $A_{\text{seg}} = \frac{1}{2}(100)\left(\frac{\pi}{3} - \sin\left(\frac{\pi}{3}\right)\right)$.
[M1] Exact form: $50\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) \approx 50(1.0472 - 0.86603) = 50(0.18117)$.
[A1] Calculate: $A_{\text{seg}} \approx \mathbf{9.06\text{ cm}^2}$.
Question 7 Tier 2: Intermediate 4 Marks

Solve the trigonometric equation for $0 \le \theta \le 2\pi$: $2\cos \theta + 1 = 0$.

View Full Method Markscheme
[M1] Isolate $\cos \theta$: $2\cos \theta = -1 \implies \cos \theta = -\frac{1}{2}$.
[M1] Find reference angle: $\cos(\theta_{\text{ref}}) = \frac{1}{2} \implies \theta_{\text{ref}} = \frac{\pi}{3}$.
[M1] Cosine is negative in Quadrants 2 and 3.
[A1] Q2 solution: $\pi - \frac{\pi}{3} = \mathbf{\frac{2\pi}{3}}$. Q3 solution: $\pi + \frac{\pi}{3} = \mathbf{\frac{4\pi}{3}}$. Solutions: $\mathbf{\theta = \frac{2\pi}{3}, \; \frac{4\pi}{3}}$.
Question 8 Tier 3: Extended 5 Marks

A pendulum of length $80\text{ cm}$ swings through an angle of $15^\circ$.
(a) Convert $15^\circ$ to exact radians.
(b) Calculate the distance travelled by the pendulum bob in one complete swing (from one extreme to the other).
(c) Find the vertical rise of the bob from its lowest position.

View Full Method Markscheme
[M1] Part (a): $15^\circ = 15 \times \frac{\pi}{180} = \mathbf{\frac{\pi}{12}\text{ rad}}$.
[M1] Part (b): Arc length $s = r\theta = 80 \times \frac{\pi}{12} = \mathbf{\frac{20\pi}{3}\text{ cm} \approx 20.9\text{ cm}}$.
[M1] Part (c): Half-swing angle is $7.5^\circ$. Distance from pivot to horizontal line through bob is $80 \cos(7.5^\circ)$.
[M1] Vertical rise $\Delta h = 80 - 80 \cos(7.5^\circ) = 80(1 - \cos(7.5^\circ))$.
[A1] $\cos(7.5^\circ) \approx 0.99144 \implies \Delta h \approx 80(0.008555) \approx \mathbf{0.684\text{ cm}}$ (or $6.84\text{ mm}$).
Question 9 Tier 3: Extended 5 Marks

Two pulleys of radii $18\text{ cm}$ and $6\text{ cm}$ have centers separated by $24\text{ cm}$. An external belt wraps tautly around both pulleys.
(a) Show that the angle of lap on the larger pulley is $\frac{4\pi}{3}\text{ radians}$.
(b) Calculate the total length of the belt.

View Full Method Markscheme
[M1] Draw right triangle with hypotenuse $d = 24$ and leg $(R - r) = 18 - 6 = 12\text{ cm}$.
[M1] Angle $\cos \alpha = \frac{R - r}{d} = \frac{12}{24} = \frac{1}{2} \implies \alpha = \frac{\pi}{3} = 60^\circ$.
[M1] Contact angle on large pulley: $2\pi - 2\alpha = 2\pi - \frac{2\pi}{3} = \mathbf{\frac{4\pi}{3}\text{ rad}}$. Contact angle on small pulley: $2\alpha = \frac{2\pi}{3}\text{ rad}$. [Proved]
[M1] Straight segment length by Pythagoras: $L_{\text{straight}} = \sqrt{24^2 - 12^2} = \sqrt{576 - 144} = \sqrt{432} = 12\sqrt{3}\text{ cm}$ (two such segments).
[A1] Total length: $2(12\sqrt{3}) + 18\left(\frac{4\pi}{3}\right) + 6\left(\frac{2\pi}{3}\right) = 24\sqrt{3} + 24\pi + 4\pi = \mathbf{24\sqrt{3} + 28\pi\text{ cm}} \approx 129.5\text{ cm}$.
Question 10 Tier 3: Extended 5 Marks

Find all solutions in radians to the trigonometric equation for $0 \le \theta < 2\pi$: $$2\sin^2 \theta - 3\sin \theta + 1 = 0$$

View Full Method Markscheme
[M1] Treat as a quadratic in $\sin \theta$: $(2\sin \theta - 1)(\sin \theta - 1) = 0$.
[M1] Case 1: $\sin \theta = \frac{1}{2}$. Reference angle is $\frac{\pi}{6}$. Positive in Q1 and Q2.
[A1] Solutions for Case 1: $\mathbf{\theta = \frac{\pi}{6}, \; \frac{5\pi}{6}}$.
[M1] Case 2: $\sin \theta = 1$. The only point on the unit circle with $y = 1$ is the top vertex.
[A1] Solution for Case 2: $\mathbf{\theta = \frac{\pi}{2}}$. Complete set of solutions: $\mathbf{\left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\right\}}$.

Radians & Unit Circle Formulas

  • Radian-Degree Identity: $\pi\text{ rad} = 180^\circ$
  • Arc Length: $s = r\theta \quad (\theta \text{ in radians})$
  • Sector Area: $A = \dfrac{1}{2}r^2\theta \quad (\theta \text{ in radians})$
  • Segment Area: $A = \dfrac{1}{2}r^2(\theta - \sin \theta)$
  • Unit Circle: $(x, y) = (\cos \theta, \sin \theta)$
  • CAST Rule: Q1 All, Q2 Sin, Q3 Tan, Q4 Cos

Top Common MYP Pitfalls

1. Using Degrees in $s = r\theta$

Entering $\theta$ in degrees into $s = r\theta$ or $A = \frac{1}{2}r^2\theta$. These formulas ONLY work when $\theta$ is in radians!

2. Calculator Mode Mismatch in Segment Area

In $\theta - \sin \theta$, keeping calculator in degree mode when computing $\sin \theta$. The entire expression assumes radians!

3. Missing the Second Quadrant Solution

For equations like $\sin \theta = \frac{1}{2}$, stopping after finding $\frac{\pi}{6}$ and forgetting $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

4. Reference Angle to $y$-Axis

Measuring reference angle from the $y$-axis. The reference angle is ALWAYS measured to the horizontal $x$-axis!

Interactive Mini Test: Radians & Unit Circle

Answer the questions to see your score and feedback.

Question 1 of 5 Degree to Radian Conversion

What is $240^\circ$ expressed in exact radians?

Question 2 of 5 Arc Length Calculation

What is the arc length of a circle of radius $12\text{ cm}$ subtended by a central angle of $\dfrac{\pi}{4}\text{ radians}$?

Question 3 of 5 Sector Area

What is the area of a sector with radius $6\text{ cm}$ and central angle $\dfrac{2\pi}{3}\text{ radians}$?

Question 4 of 5 CAST Diagram Signs

In Quadrant 4, which of the three primary trigonometric ratios is positive?

Question 5 of 5 Unit Circle Value

What is the exact value of $\cos(\pi)$?