Foundations — Radian Measure, Standard Position, and the Unit Circle

This article is also available in 简体中文, 繁體中文.

0. Why start with angle measure?

In trigonometry, everything is built on the idea of turning—how far a ray rotates about a point. To compute lengths, areas, and later define sine/cosine cleanly, we need a natural unit for angles. Degrees are familiar for navigation and everyday life, but radians are the unit that makes geometry and calculus “just work”—they tie angles directly to arc length. This lecture formalizes that connection and equips you with all the foundational tools you’ll use across the rest of Chapter 1.

1. The radian: the natural unit of angle

1.1 Central angles and arc length

Consider a circle of radius $r$ and a central angle $\theta$ (measured at the center). The arc cut out by this angle has length $s$. The radian is defined so that the angle is the ratio of arc length to radius:
$$
\boxed{\ \theta=\dfrac{s}{r}\ }\quad\text{(radians)}.
$$

Why this is beautiful. No arbitrary scale is imposed—the unit is “built in” to the geometry. If $r=1$ (the unit circle), then $\theta=s$: the angle equals the arc length. That one-to-one link is what makes radian measure indispensable.

Dimensional check. $s$ and $r$ both have units of length, so $\theta$ is dimensionless. In practice we still write “rad” to remind ourselves which unit we’re using.

---

1.2 Full turn and the radian–degree bridge

A full revolution corresponds to the full circumference $2\pi r$. Plugging $s=2\pi r$ into $\theta=s/r$ gives
$$
\text{one full turn}=2\pi\ \text{radians}.
$$
Since one full turn is also $360^\circ$, we get the conversion facts:
$$
\boxed{\ 180^\circ=\pi\ \text{rad}\ }\quad\Longleftrightarrow\quad
\boxed{\ 1^\circ=\frac{\pi}{180}\ \text{rad},\ \ 1\ \text{rad}=\frac{180}{\pi}^\circ\ (\approx57.3^\circ). }
$$

Quick conversions (know these cold).
$$
\begin{array}{c|cccccccccc}
\text{Degrees} & 0^\circ&30^\circ&45^\circ&60^\circ&90^\circ&120^\circ&135^\circ&150^\circ&180^\circ&360^\circ\\hline
\text{Radians} & 0&\frac{\pi}{6}&\frac{\pi}{4}&\frac{\pi}{3}&\frac{\pi}{2}&\frac{2\pi}{3}&\frac{3\pi}{4}&\frac{5\pi}{6}&\pi&2\pi
\end{array}
$$

Technique tip. To convert degrees $\to$ radians, multiply by $\pi/180$. To convert radians $\to$ degrees, multiply by $180/\pi$. Keep fractions in lowest terms; it helps downstream when you place angles on the circle.

---

2. Standard position and sign conventions

2.1 Definition (standard position)

An angle is in standard position if its vertex is at the origin and its initial ray lies on the positive $x$-axis. Rotating counterclockwise gives positive angles; rotating clockwise gives negative angles.

• Example: $+\tfrac{\pi}{3}$ is a $60^\circ$ counterclockwise rotation; $-\tfrac{\pi}{2}$ is a $90^\circ$ clockwise rotation.

Beyond a single turn. There’s no upper bound. You can have $\theta=5\pi$ (two and a half turns) or $\theta=-9\pi/4$ (clockwise > two turns). We will default to radians unless explicitly stated otherwise.

---

2.2 Coterminal angles and reduction modulo $2\pi$

Two angles are coterminal if they have the same terminal ray when drawn in standard position. This happens exactly when they differ by a multiple of $2\pi$:
$$
\theta \ \text{and}\ \theta+2\pi k\quad (k\in\mathbb Z)\ \text{are coterminal}.
$$
This motivates angle reduction: given any $\theta$, we can replace it by an equivalent angle in the principal interval $(-\pi,\pi]$ or $[0,2\pi)$ by adding/subtracting $2\pi$ repeatedly.

• Example: Reduce $\theta=\dfrac{29\pi}{6}$ to $[0,2\pi)$.
  $$
  \frac{29\pi}{6}-2\pi=\frac{29\pi}{6}-\frac{12\pi}{6}=\frac{17\pi}{6}>2\pi,\quad
  \frac{17\pi}{6}-2\pi=\frac{17\pi}{6}-\frac{12\pi}{6}=\frac{5\pi}{6}\in[0,2\pi).
  $$
  So $\dfrac{29\pi}{6}$ is coterminal with $\dfrac{5\pi}{6}$.

---

3. Arc length and sector area

A central angle $\theta$ (in radians) cuts out an arc of length
$$
\boxed{\ s=r\theta\ } \qquad(\theta\ \text{in radians only!})
$$
and a sector (pie-slice) of area
$$
\boxed{\ A=\tfrac12 r^2\theta\ } \qquad(\theta\ \text{in radians only!})
$$
These are linear (for $s$) and quadratic (for $A$) in the radius—consistent with dimensional reasoning.

Derivation (sector area). A full circle area is $\pi r^2$, corresponding to angle $2\pi$. Proportionally,
$$
\frac{A}{\pi r^2}=\frac{\theta}{2\pi}\quad\Rightarrow\quad A=\frac{\theta}{2\pi},\pi r^2=\frac12 r^2\theta.
$$

Unit circle view. When $r=1$, these collapse to:
$$
s=\theta,\qquad A=\tfrac12\theta.
$$
So on the unit circle, the number $\theta$ is the arc length; sectors have area equal to “half the angle.”

---

4. Degrees vs. radians in practice

Golden rule. Whenever you use $s=r\theta$ or $A=\tfrac12 r^2\theta$, $\theta$ must be in radians. If your angle is given in degrees, convert first.

• Right: $s = 10\cdot\big(60^\circ\cdot \tfrac{\pi}{180}\big)=10\cdot\tfrac{\pi}{3}$.
• Wrong: $s = 10\cdot 60$ (treated $60^\circ$ as a pure 60; this is off by a factor of $\pi/180$).

Mental benchmarks.

• $30^\circ=\pi/6\approx 0.524$ rad
• $45^\circ=\pi/4\approx 0.785$ rad
• $60^\circ=\pi/3\approx 1.047$ rad
• $90^\circ=\pi/2\approx 1.571$ rad

These help you sanity-check numeric outputs.

---

5. Worked examples (step-by-step)

Example 1 — Convert angles between degrees and radians

(a) $315^\circ$ to radians.
Multiply by $\pi/180$:
$$
315^\circ=\frac{315\pi}{180}=\frac{7\pi}{4}.
$$

(b) $-\dfrac{11\pi}{6}$ to degrees.
Multiply by $180/\pi$:
$$
-\frac{11\pi}{6}\cdot\frac{180}{\pi}=-\frac{11\cdot 180}{6}=-330^\circ.
$$

(c) $\dfrac{23\pi}{3}$ reduced to $[0,2\pi)$ in radians (also report degrees).
Subtract $2\pi=\dfrac{6\pi}{3}$ repeatedly:
$$
\frac{23\pi}{3}-\frac{18\pi}{3}=\frac{5\pi}{3}\in[0,2\pi).
$$
In degrees, $\frac{5\pi}{3}\cdot\frac{180}{\pi}=300^\circ$.

---

Example 2 — Arc length and sector area

A circle has radius $r=12\ \mathrm{cm}$. Find the arc length $s$ and sector area $A$ cut off by $\theta=150^\circ$.

1. Convert $\theta$ to radians: $150^\circ=\dfrac{5\pi}{6}$.
2. Arc length: $s=r\theta=12\cdot\dfrac{5\pi}{6}=10\pi\ \mathrm{cm}$.
3. Sector area: $A=\tfrac12 r^2\theta=\tfrac12\cdot 144\cdot \dfrac{5\pi}{6}=12\cdot \dfrac{5\pi}{6}=10\pi\ \mathrm{cm}^2$.

Notice: Coincidentally here $s$ and $A$ both equal $10\pi$ but with different units; this is not a general rule. It occurred because $r=12$ and $\theta=5\pi/6$ happened to make $r\theta=\tfrac12 r^2\theta$ numerically equal.

---

Example 3 — Negative and large angles, standard position

Sketch $\theta=-\dfrac{7\pi}{4}$ and $\theta=\dfrac{19\pi}{6}$ in standard position. Identify a coterminal angle in $[0,2\pi)$.

• For $\theta=-\dfrac{7\pi}{4}$: Add $2\pi=\dfrac{8\pi}{4}$ once:
  $$
  -\frac{7\pi}{4}+2\pi=\frac{\pi}{4}\quad(\text{in }[0,2\pi)).
  $$
  So the terminal side coincides with $\pi/4$ (first quadrant). Draw a $45^\circ$ clockwise rotation from the positive $x$-axis (since the original is negative).

• For $\theta=\dfrac{19\pi}{6}$: Subtract $2\pi=\dfrac{12\pi}{6}$:
  $$
  \frac{19\pi}{6}-2\pi=\frac{7\pi}{6}\quad(\text{in }[0,2\pi)).
  $$
  So it lands at $210^\circ$ (third quadrant).

---

Example 4 — Unit circle interpretation

Let $r=1$ and $\theta=\dfrac{3\pi}{2}$.

• Arc length along the unit circle: $s=\theta=\dfrac{3\pi}{2}$.
• Sector area: $A=\tfrac12\theta=\dfrac{3\pi}{4}$.
  Interpretation: a $270^\circ$ sweep covers three-quarters of the circle, so its area is $ \tfrac{3}{4}\cdot \pi(1)^2 = \tfrac{3\pi}{4}$, consistent with $A=\tfrac12\theta$.

---

Example 5 — Mixed units sanity check

A wheel of radius $0.35,\mathrm{m}$ spins through $1.8$ radians.

• $s=r\theta=0.35\cdot 1.8=0.63\ \mathrm{m}$.
• If someone reported $s=63\ \mathrm{m}$, the result is absurd—unit check (and the scale of numbers) immediately flags the error.

---

6. Reference angles and compass of the circle (preparation)

Although we formally define trigonometric functions in the next part, it is already helpful to know the standard angles around the circle and their locations. These will be your “map pins” when evaluating or sketching later.

• Quadrantal angles (on axes): $0,\ \tfrac{\pi}{2},\ \pi,\ \tfrac{3\pi}{2},\ 2\pi$.
• Special angles in the first quadrant: $\tfrac{\pi}{6}\ (30^\circ),\ \tfrac{\pi}{4}\ (45^\circ),\ \tfrac{\pi}{3}\ (60^\circ)$. Their reflections/rotations generate the common points in other quadrants.

Reference angle. For any non-quadrantal $\theta$, the reference angle $\theta_{\text{ref}}\in(0,\tfrac{\pi}{2})$ is the acute angle between the terminal side of $\theta$ and the nearest $x$-axis. Angle reduction often proceeds by finding $\theta_{\text{ref}}$ after locating the correct quadrant.

• Example: $\theta=\dfrac{11\pi}{6}$ sits in QIV and has $\theta_{\text{ref}}=2\pi-\dfrac{11\pi}{6}=\dfrac{\pi}{6}$.

(We will use $\theta_{\text{ref}}$ heavily when we start evaluating trig functions.)

---

7. Common pitfalls and how to avoid them

1. Forgetting radians in $s=r\theta$ or $A=\tfrac12r^2\theta$.
   Always convert degrees first; highlight $\theta$ in your notes and label the unit.

2. Not reducing angles before sketching.
   A quick subtraction of $2\pi$ can reveal the quadrant immediately.

3. Sign mistakes with negative angles.
   Remember: clockwise is negative; a negative angle’s terminal side may coincide with a positive coterminal angle.

4. Mixing numeric approximations and symbols haphazardly.
   Keep $\pi$ as a symbol unless the question requests a decimal.

5. Dimensional sense.
   Arc lengths scale linearly with $r$, areas scale with $r^2$. If increasing $r$ by a factor of 10 leaves your $A$ unchanged, something is wrong.

---

8. Short practice (with answers)

Try these before peeking at the solutions.

Q1. Convert to radians:
(a) $210^\circ$ (b) $-75^\circ$ (c) $810^\circ$.

Q2. Convert to degrees:
(a) $\dfrac{7\pi}{12}$ (b) $-\dfrac{5\pi}{3}$ (c) $\dfrac{25\pi}{4}$ reduced to $[0,360^\circ)$.

Q3. A sector has radius $r=8$ cm and angle $\theta=\dfrac{3\pi}{5}$. Find $s$ and $A$.

Q4. Find a coterminal angle of $-\dfrac{23\pi}{6}$ in $[0,2\pi)$. Sketch in standard position.

Q5. A wheel of radius $0.5$ m rotates through $225^\circ$. Find the distance a point on the rim travels.

Answers.

• A1. (a) $210^\circ=\dfrac{210\pi}{180}=\dfrac{7\pi}{6}$.
  (b) $-75^\circ=-\dfrac{75\pi}{180}=-\dfrac{5\pi}{12}$.
  (c) $810^\circ=\dfrac{810\pi}{180}=\dfrac{9\pi}{2}$.

• A2. (a) $\dfrac{7\pi}{12}\cdot\dfrac{180}{\pi}=105^\circ$.
  (b) $-\dfrac{5\pi}{3}\cdot\dfrac{180}{\pi}=-300^\circ$.
  (c) $\dfrac{25\pi}{4}\cdot\dfrac{180}{\pi}=1125^\circ$. Reduce modulo $360^\circ$: $1125-720=405$, $405-360=45^\circ$.

• A3. $s=r\theta=8\cdot \dfrac{3\pi}{5}=\dfrac{24\pi}{5}\ \mathrm{cm}$.
  $A=\tfrac12 r^2\theta=\tfrac12\cdot 64\cdot \dfrac{3\pi}{5}=\dfrac{96\pi}{5}\ \mathrm{cm}^2$.

• A4. Add $2\pi=\dfrac{12\pi}{6}$ enough times:
  $-\dfrac{23\pi}{6}+2\pi=-\dfrac{11\pi}{6}$ (still negative), add again: $-\dfrac{11\pi}{6}+2\pi=\dfrac{\pi}{6}$. So $\dfrac{\pi}{6}$ (Q I).

• A5. Convert $225^\circ=\dfrac{5\pi}{4}$. Then $s=r\theta=0.5\cdot \dfrac{5\pi}{4}=\dfrac{5\pi}{8}\ \mathrm{m}$.

---

9. Micro-checkpoints (use during problem sets)

• C1. Did I label my angle’s unit (° or rad) at the start?
• C2. If I used $s=r\theta$ or $A=\tfrac12 r^2\theta$, did I ensure $\theta$ is in radians?
• C3. Have I reduced my angle to $[0,2\pi)$ (or $(-\pi,\pi]$) before sketching?
• C4. Does the magnitude of my answer make sense (units, size, scaling with $r$)?

---

10. One-page summary (carry sheet)

• Radian definition: $\displaystyle \theta=\frac{s}{r}$.
• Full turn: $2\pi\ \text{rad}=360^\circ$.
• Conversions: $1^\circ=\dfrac{\pi}{180}\ \text{rad},\quad 1\ \text{rad}=\dfrac{180}{\pi}^\circ$.
• Arc length: $\displaystyle s=r\theta$ (radians).
• Sector area: $\displaystyle A=\frac12 r^2\theta$ (radians).
• Standard position: vertex at origin, initial ray on $+x$, CCW positive, CW negative.
• Coterminal: $\theta\sim\theta+2\pi k$.
• Principal intervals: $[0,2\pi)$ or $(-\pi,\pi]$ for quick location.
• Unit circle shortcuts: $\pi/6, \pi/4, \pi/3$ and their multiples; know the quadrantal angles.

---

11. Where we go next

With angle measure solid, we will define the six trigonometric functions via the unit circle and right triangles, develop their core identities, and start solving basic equations. Keep this foundation handy—every identity and graph you’ll see sits on top of radians, standard position, and the geometry of the circle.