← Module 4 Visual Lab Pre-Calculus Student Support
Mathematical Architects · Pre-Calculus · Foundations

Get Ready: Module 4 — Trig Identities, Equations, Laws & Vectors

Everyone starts somewhere. Before we solve any triangle or add any vector, we'll re-pour the floor. This page rebuilds the handful of skills Module 4 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 4 takes the trig you already know and puts it to work on any triangle and on directed quantities (vectors). If the unit circle, fractions, the Pythagorean theorem, or SOH-CAH-TOA 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

Unit-circle exact values

What it is: the exact sine and cosine of the "famous" angles — \(0,\ \tfrac{\pi}{6},\ \tfrac{\pi}{4},\ \tfrac{\pi}{3},\ \tfrac{\pi}{2}\) (that's \(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\)) and their reflections around the circle. On the unit circle, a point is \((\cos\theta,\ \sin\theta)\). Why you need it: verifying identities and solving trig equations both rely on recognizing these values instantly — e.g. knowing that \(\sin\theta = \tfrac12\) happens at \(\theta = \tfrac{\pi}{6}\).

Worked example — read \(\cos\tfrac{\pi}{6}\) and \(\sin\tfrac{\pi}{6}\)
  1. Locate the angle. \(\tfrac{\pi}{6} = 30^\circ\) sits just above the positive \(x\)-axis.
  2. Use the 30-60-90 values. The coordinates are \(\left(\tfrac{\sqrt{3}}{2},\ \tfrac{1}{2}\right)\).
  3. Match \(x\) to cosine, \(y\) to sine. So \(\cos\tfrac{\pi}{6} = \tfrac{\sqrt{3}}{2}\) and \(\sin\tfrac{\pi}{6} = \tfrac12\).
  4. Get tangent for free. \(\tan\tfrac{\pi}{6} = \dfrac{\sin}{\cos} = \dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}\).

Memory aid: at \(30^\circ, 45^\circ, 60^\circ\) the sines are \(\tfrac{\sqrt{1}}{2},\ \tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{3}}{2}\) — just \(\sqrt{1}, \sqrt{2}, \sqrt{3}\) over 2.

Try it 1 — What are \(\cos\tfrac{\pi}{4}\) and \(\sin\tfrac{\pi}{4}\)?
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At \(45^\circ\) the point is \(\left(\tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{2}}{2}\right)\), so \(\cos\tfrac{\pi}{4} = \sin\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2}\). (They're equal — the \(45^\circ\) line is the diagonal.)

Try it 2 — For which first-quadrant angle is \(\cos\theta = \tfrac12\)?
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\(\cos\theta = \tfrac12\) at \(\theta = \tfrac{\pi}{3} = 60^\circ\). (There the point is \(\left(\tfrac12,\ \tfrac{\sqrt{3}}{2}\right)\).)

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

Algebra with fractions

What it is: adding, simplifying, and cross-multiplying fractions — including ones with variables. Adding needs a common denominator; a proportion \(\tfrac{p}{q} = \tfrac{r}{s}\) is solved by cross-multiplying to \(ps = qr\). Why you need it: verifying identities is fraction algebra in disguise (combining \(\sin/\cos\) terms), and the Law of Sines is a proportion you solve by cross-multiplying.

Worked example — solve the proportion \(\dfrac{\sin B}{9} = \dfrac{\sin 35^\circ}{7}\)
  1. Cross-multiply. \(7\sin B = 9\sin 35^\circ\).
  2. Isolate the unknown. \(\sin B = \dfrac{9\sin 35^\circ}{7}\).
  3. Evaluate. \(\sin 35^\circ \approx 0.5736\), so \(\sin B \approx \dfrac{9(0.5736)}{7} \approx 0.7373\).
  4. Undo the sine. \(B = \sin^{-1}(0.7373) \approx 47.5^\circ\). (This is exactly a Law-of-Sines step!)

To add fractions instead: \(\dfrac{1}{\cos\theta} + \dfrac{\sin\theta}{\cos\theta} = \dfrac{1 + \sin\theta}{\cos\theta}\) — same denominator, so just add the tops.

Try it 1 — Solve \(\dfrac{x}{12} = \dfrac{3}{8}\).
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Cross-multiply: \(8x = 36\), so \(x = \tfrac{36}{8} = \tfrac{9}{2} = 4.5\).

Try it 2 — Combine \(\dfrac{\cos\theta}{\sin\theta} + \dfrac{1}{\sin\theta}\) into a single fraction.
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Same denominator \(\sin\theta\), so add the numerators: \(\dfrac{\cos\theta + 1}{\sin\theta}\).

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

The Pythagorean theorem

What it is: in a right triangle, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse (the side across from the right angle). Why you need it: it's the seed of two big ideas this module — the magnitude of a vector \(\langle x, y\rangle\) is \(\sqrt{x^2 + y^2}\) (Pythagoras), and the Law of Cosines is Pythagoras generalized to any triangle (with a \(-2ab\cos C\) correction term).

Worked example — find the missing side of a right triangle with legs \(3\) and \(4\)
  1. Identify the parts. The two legs are \(a = 3\) and \(b = 4\); \(c\) is the hypotenuse.
  2. Square the legs. \(3^2 + 4^2 = 9 + 16 = 25\).
  3. Take the square root. \(c = \sqrt{25} = 5\).
  4. So the hypotenuse is \(5\). (The famous 3-4-5 right triangle.)

Same arithmetic gives a vector's length: \(\langle 3, 4\rangle\) has magnitude \(\sqrt{3^2 + 4^2} = 5\).

Try it 1 — A right triangle has legs \(5\) and \(12\). Find the hypotenuse.
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\(5^2 + 12^2 = 25 + 144 = 169\), so \(c = \sqrt{169} = 13\). (The 5-12-13 triangle.)

Try it 2 — What is the magnitude of the vector \(\langle 7, 5\rangle\)?
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\(\sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.60\). (It need not be a whole number.)

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

Basic right-triangle trig (SOH-CAH-TOA)

What it is: in a right triangle, the three ratios — \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\). The mnemonic is SOH-CAH-TOA. Why you need it: the Laws of Sines and Cosines are the "any triangle" upgrade of these ratios, and you'll use \(\tan^{-1}\) to recover a vector's direction angle from its components.

Worked example — find angle \(\theta\) when opposite \(= 5\), adjacent \(= 7\)
  1. Pick the ratio with the sides you have. Opposite and adjacent → use tangent: \(\tan\theta = \dfrac{\text{opp}}{\text{adj}} = \dfrac{5}{7}\).
  2. Undo the tangent. \(\theta = \tan^{-1}\!\left(\tfrac{5}{7}\right)\).
  3. Evaluate. \(\theta \approx 35.5^\circ\).
  4. So \(\theta \approx 35.5^\circ\). This is exactly how the lab reports a vector's direction.

Choosing the ratio: write down which two sides you know (opp / adj / hyp), then pick the rule that uses both.

Try it 1 — In a right triangle, the side opposite \(\theta\) is \(6\) and the hypotenuse is \(10\). Find \(\sin\theta\) and \(\theta\).
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\(\sin\theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{6}{10} = 0.6\), so \(\theta = \sin^{-1}(0.6) \approx 36.9^\circ\).

Try it 2 — Find the direction angle of the vector \(\langle 4, 4\rangle\).
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\(\theta = \tan^{-1}\!\left(\tfrac{4}{4}\right) = \tan^{-1}(1) = 45^\circ\). (Equal components → the \(45^\circ\) diagonal.)

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. What are \(\cos\tfrac{\pi}{3}\) and \(\sin\tfrac{\pi}{3}\)?
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    At \(60^\circ\): \(\cos\tfrac{\pi}{3} = \tfrac12\), \(\sin\tfrac{\pi}{3} = \tfrac{\sqrt{3}}{2}\).

  2. Solve the proportion \(\dfrac{\sin C}{10} = \dfrac{\sin 40^\circ}{8}\) for \(\sin C\).
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    Cross-multiply: \(8\sin C = 10\sin 40^\circ\), so \(\sin C = \dfrac{10\sin 40^\circ}{8} \approx 0.803\).

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

  4. Find the magnitude of the vector \(\langle 6, 8\rangle\).
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    \(\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\).

  5. In a right triangle, opposite \(= 3\), adjacent \(= 4\). Find \(\theta\).
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    \(\theta = \tan^{-1}\!\left(\tfrac{3}{4}\right) \approx 36.9^\circ\).

  6. Combine \(\dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta}\) into one fraction.
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    Common denominator already shared: \(\dfrac{1 - \sin\theta}{\cos\theta}\).

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 every one of these ideas move at once.

Module 4 Overview & Visual Lab Jump to the Visual Lab Student Support