← Module 3 Visual Lab Pre-Calculus Student Support
Mathematical Architects · Pre-Calculus · Foundations

Get Ready: Module 3 — Trigonometric Functions & the Unit Circle

Everyone starts somewhere. Before we wrap a triangle around a circle, we'll re-pour the floor. This page rebuilds the handful of skills Module 3 leans on — clearly, one step at a time, with nothing assumed. Go at your own pace. There's no rush and no wrong place to begin.

4 Foundation Skills Worked Examples Self-Paced

Module 3 takes the right triangle you already know and bends it onto a circle, so the same ratios start to repeat. If SOH-CAH-TOA, the Pythagorean theorem, the coordinate plane, or fractions of \(\pi\) feel fuzzy right now, that's completely okay — we'll firm them up together below, then you'll be ready for the lab.

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Skills to build first

Four short skills. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.

1 Skill

Right-triangle trig (SOH-CAH-TOA)

What it is: in a right triangle, the three ratios sine, cosine, and tangent compare two sides relative to an acute angle \(\theta\): \(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\). Why you need it: the unit circle is just a right triangle with hypotenuse \(1\). Once the hypotenuse is \(1\), \(\sin\theta\) is the opposite side and \(\cos\theta\) is the adjacent side — which become the \(y\) and \(x\) coordinates of the point on the circle.

Worked example — a 3–4–5 right triangle
  1. Label the sides for the angle \(\theta\) whose opposite leg is \(3\), adjacent leg is \(4\), and hypotenuse is \(5\).
  2. SOH: \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{3}{5} = 0.6\).
  3. CAH: \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{4}{5} = 0.8\).
  4. TOA: \(\tan\theta = \dfrac{\text{opp}}{\text{adj}} = \dfrac{3}{4} = 0.75\).

The phrase SOH-CAH-TOA just packs those three definitions into one memorable word.

Try it 1 — In a right triangle, the side opposite \(\theta\) is \(5\), the adjacent side is \(12\), and the hypotenuse is \(13\). Find \(\sin\theta\) and \(\cos\theta\).
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\(\sin\theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{5}{13}\); \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{12}{13}\).

Try it 2 — For the same triangle, find \(\tan\theta\).
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\(\tan\theta = \dfrac{\text{opp}}{\text{adj}} = \dfrac{5}{12}\). (Notice \(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{5/13}{12/13} = \dfrac{5}{12}\) too.)

Practice more (free) Khan Academy → IXL →
2 Skill

The Pythagorean theorem

What it is: in a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^2 + b^2 = c^2\). Why you need it: it's the reason the unit circle has the equation \(x^2 + y^2 = 1\), and it's the source of the most important trig identity you'll meet next module: \(\sin^2\theta + \cos^2\theta = 1\). It's also how you find a missing coordinate from the other one.

Worked example — find the hypotenuse for legs \(6\) and \(8\)
  1. Square the legs. \(a^2 + b^2 = 6^2 + 8^2 = 36 + 64 = 100\).
  2. That equals \(c^2\). So \(c^2 = 100\).
  3. Take the positive square root (a length): \(c = \sqrt{100} = 10\).

On the unit circle the hypotenuse is fixed at \(1\), so the relationship becomes \(x^2 + y^2 = 1\): the coordinates are never independent.

Try it 1 — A right triangle has legs \(9\) and \(12\). Find the hypotenuse.
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\(9^2 + 12^2 = 81 + 144 = 225\), so \(c = \sqrt{225} = 15\).

Try it 2 — A point on the unit circle has \(x = \tfrac{1}{2}\). If it's in Quadrant I, find \(y\). (Hint: \(x^2 + y^2 = 1\).)
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\(\big(\tfrac12\big)^2 + y^2 = 1 \Rightarrow \tfrac14 + y^2 = 1 \Rightarrow y^2 = \tfrac34\). In Quadrant I, \(y > 0\), so \(y = \dfrac{\sqrt3}{2}\). (That's the point at \(60^\circ\)!)

Practice more (free) Khan Academy → IXL →
3 Skill

The coordinate plane & quadrants

What it is: the grid where every point has an address \((x, y)\) — \(x\) tells you left/right, \(y\) tells you up/down — split into four quadrants by the axes. Why you need it: the unit circle lives on this plane, and the quadrant a terminal point falls in decides the sign of \(\cos\theta\) (the x-coordinate) and \(\sin\theta\) (the y-coordinate). Getting the signs right is the single most common unit-circle slip.

Worked example — signs by quadrant
  1. Quadrant I (right & up): \(x > 0,\ y > 0\) → both \(\cos\) and \(\sin\) positive.
  2. Quadrant II (left & up): \(x < 0,\ y > 0\) → \(\cos\) negative, \(\sin\) positive.
  3. Quadrant III (left & down): \(x < 0,\ y < 0\) → both negative.
  4. Quadrant IV (right & down): \(x > 0,\ y < 0\) → \(\cos\) positive, \(\sin\) negative.

A classic mnemonic: "All Students Take Calculus" — in QI All are positive, QII Sine, QIII Tangent, QIV Cosine.

Try it 1 — A terminal point is in Quadrant III. Are \(\cos\theta\) and \(\sin\theta\) positive or negative?
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Quadrant III is left and down, so \(x < 0\) and \(y < 0\): both \(\cos\theta\) and \(\sin\theta\) are negative.

Try it 2 — If \(\cos\theta > 0\) and \(\sin\theta < 0\), which quadrant is the angle in?
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Positive x and negative y → right and down → Quadrant IV.

Practice more (free) Khan Academy → IXL →
4 Skill

Fractions of \(\pi\)

What it is: comfort with \(\pi\) as a number (about \(3.14\)) that you can slice into fractions like \(\tfrac{\pi}{2}\), \(\tfrac{\pi}{4}\), \(\tfrac{2\pi}{3}\) — and adding, comparing, and reducing them. Why you need it: radian angles are fractions of \(\pi\). A half-turn is \(\pi\), a quarter-turn is \(\tfrac{\pi}{2}\), and the special angles you'll memorize are all \(\pi\) over something. Converting \(180^\circ = \pi\) and back is the whole skill of degree↔radian.

Worked example — convert and place \(\tfrac{\pi}{4}\) and \(\tfrac{3\pi}{4}\)
  1. Anchor on \(180^\circ = \pi\). So \(\tfrac{\pi}{4} = \tfrac{180^\circ}{4} = 45^\circ\).
  2. Scale up. \(\tfrac{3\pi}{4} = 3 \cdot \tfrac{\pi}{4} = 3 \cdot 45^\circ = 135^\circ\).
  3. Compare them. Since \(45^\circ < 135^\circ\), the angle \(\tfrac{\pi}{4}\) is smaller — just under halfway to a right angle, while \(\tfrac{3\pi}{4}\) is past it.
  4. Add fractions of \(\pi\) like any fractions. \(\tfrac{\pi}{4} + \tfrac{\pi}{2} = \tfrac{\pi}{4} + \tfrac{2\pi}{4} = \tfrac{3\pi}{4}\) (common denominator first).
Try it 1 — Convert \(\tfrac{\pi}{6}\) and \(\tfrac{\pi}{3}\) to degrees.
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\(\tfrac{\pi}{6} = \tfrac{180^\circ}{6} = 30^\circ\); \(\tfrac{\pi}{3} = \tfrac{180^\circ}{3} = 60^\circ\).

Try it 2 — Add \(\tfrac{\pi}{3} + \tfrac{\pi}{6}\), then convert the result to degrees.
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Common denominator \(6\): \(\tfrac{2\pi}{6} + \tfrac{\pi}{6} = \tfrac{3\pi}{6} = \tfrac{\pi}{2}\). That's \(90^\circ\) — a quarter turn.

Practice more (free) Khan Academy → IXL →
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Quick Readiness Check

Six short questions spanning all four skills. Try each one first, then reveal the answer to check yourself.

  1. In a right triangle, the side opposite \(\theta\) is \(7\) and the hypotenuse is \(25\). What is \(\sin\theta\)?
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    \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{7}{25}\).

  2. A right triangle has legs \(8\) and \(15\). Find the hypotenuse.
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    \(8^2 + 15^2 = 64 + 225 = 289\), so \(c = \sqrt{289} = 17\).

  3. In which quadrant is \(\cos\theta < 0\) and \(\sin\theta > 0\)?
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    Negative x, positive y → left and up → Quadrant II.

  4. Convert \(\tfrac{5\pi}{6}\) to degrees.
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    \(\tfrac{5\pi}{6} = 5 \cdot \tfrac{180^\circ}{6} = 5 \cdot 30^\circ = 150^\circ\).

  5. A point on the unit circle has \(y = \tfrac{1}{2}\) in Quadrant I. Use \(x^2 + y^2 = 1\) to find \(x\).
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    \(x^2 = 1 - \tfrac14 = \tfrac34\), and \(x > 0\) in QI, so \(x = \dfrac{\sqrt3}{2}\). (The \(30^\circ\) point.)

  6. Convert \(270^\circ\) to radians.
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    \(270^\circ \cdot \tfrac{\pi}{180^\circ} = \tfrac{270\pi}{180} = \dfrac{3\pi}{2}\).

If these feel comfortable, you're ready for the module. If one or two felt shaky, that's your cue — scroll back up to that skill, re-read the worked example, and try a couple from the free practice links. No score to hit, no clock running. Come back when it clicks.

Floor poured. When the skills above feel steady, step into the module — the Visual Lab lets you see the triangle become a circle and the circle become a wave, all at once.

Module 3 Overview & Visual Lab Jump to the Visual Lab Student Support