Module 0f — Trigonometry
Course: MAT1CJ101 — Differential Calculus (prerequisite refresher)
Status: Prerequisite background material — not part of the official 60 taught hours, not examined.
0. Why This Module Exists
Module V of the main notes (Trigonometry, Optimization, Linearization) uses trig functions inside derivatives and optimization problems, assuming you can already evaluate them, manipulate identities, and solve trig equations fluently. This module is that fluency practice, built from the unit circle up.
1. Angle Measure
1.1 Degrees and Radians
A full revolution is $360°$ or $2\pi$ radians. The conversion factor:
\[\text{radians} = \text{degrees} \times \frac{\pi}{180} \qquad \text{degrees} = \text{radians} \times \frac{180}{\pi}\]Worked Example 1: Convert $150°$ to radians, and convert $5\pi/6$ radians to degrees.
\(150° \times \frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6} \text{ radians}\) \(\frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5(180)}{6} = \frac{900}{6} = 150° \quad \text{(consistent — same angle both ways)}\)
1.2 Common Angles
| Degrees | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ | $180°$ | $270°$ | $360°$ |
|---|---|---|---|---|---|---|---|---|
| Radians | $0$ | $\pi/6$ | $\pi/4$ | $\pi/3$ | $\pi/2$ | $\pi$ | $3\pi/2$ | $2\pi$ |
Memorising this table (especially $30°, 45°, 60°, 90°$ and their radian forms) makes every later identity and equation dramatically faster to work with.
2. The Unit Circle
2.1 Definition of sin and cos via the Unit Circle
For an angle $\theta$ measured counterclockwise from the positive $x$-axis, let $(x, y)$ be the point where the terminal side meets the unit circle (radius 1, centred at origin). Define:
\[\cos\theta = x \qquad \sin\theta = y\]This is the definition that works for any real angle $\theta$, not just angles inside a right triangle ($0°$ to $90°$) — it’s what lets sin and cos be defined as genuine functions on all of $\mathbb{R}$, which is what you need once you start differentiating them.
Example 1 (Unit-circle sin/cos): The terminal side of $\theta = \pi/2$ meets the unit circle at $(0, 1)$. What are $\sin(\pi/2)$ and $\cos(\pi/2)$?
Point on the unit circle: $(x, y) = (0, 1)$. $\cos\theta = x = 0$. $\sin\theta = y = 1$. Check: $x^2 + y^2 = 0^2 + 1^2 = 1$. $\checkmark$ (genuinely on the unit circle, radius 1) So $\cos(\pi/2) = 0$, $\sin(\pi/2) = 1$.
Non-Example 1 (Not a valid unit-circle point): Can $\cos\theta = 2$ for some angle $\theta$?
Any point $(x, y)$ on the unit circle must satisfy $x^2 + y^2 = 1$, which forces $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$ (a coordinate can’t exceed the circle’s radius). Is $2$ within $[-1, 1]$? No. $\times$ $\cos\theta = 2$ is impossible for any real $\theta$ — no point on the unit circle has $x = 2$, so this “value” corresponds to no valid position on the circle at all. (Same reasoning rules out $\sin\theta = 2$, or $\cos\theta = -1.5$, etc.)
2.2 The Other Four Trig Functions
\(\tan\theta = \frac{\sin\theta}{\cos\theta} \quad \text{(undefined where } \cos\theta = 0\text{)}\) \(\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta} \quad \text{(undefined where } \sin\theta = 0\text{)}\) \(\sec\theta = \frac{1}{\cos\theta} \quad \text{(undefined where } \cos\theta = 0\text{)}\) \(\csc\theta = \frac{1}{\sin\theta} \quad \text{(undefined where } \sin\theta = 0\text{)}\)
2.3 Signs by Quadrant (ASTC / “All Students Take Calculus”)
| Quadrant | Angle range | sin | cos | tan |
|---|---|---|---|---|
| I | $0°$ to $90°$ | + | + | + |
| II | $90°$ to $180°$ | + | − | − |
| III | $180°$ to $270°$ | − | − | + |
| IV | $270°$ to $360°$ | − | + | − |
The mnemonic “All Students Take Calculus” gives, quadrant by quadrant (I,II,III,IV), which functions are positive: All, Sin, Tan, Cos.
Example 2 (ASTC sign): What is the sign of $\sin(160°)$?
$160°$ is between $90°$ and $180°$ $\to$ Quadrant II. ASTC for Quadrant II: only Sin is positive (sin $+$, cos $-$, tan $-$). So $\sin(160°)$ is POSITIVE. $\checkmark$ (matches: $160°$ is close to $180°$, still “above” the $x$-axis, so its $y$-coordinate/sine is positive)
Non-Example 2 (Common ASTC sign error): What is the sign of $\cos(160°)$?
A tempting but WRONG shortcut: “$160°$ is close to $180°$, and $\cos(180°)$ is negative, so maybe I should just check if the angle LOOKS positive-ish.” That’s not how ASTC works — always identify the quadrant first, then read straight off the table. $160°$ is in Quadrant II ($90°$ to $180°$). ASTC for Quadrant II: cos is NEGATIVE (only sin is $+$). So $\cos(160°)$ is NEGATIVE — do not assume “since sin is positive here, cos must be positive too”; each function has its own sign per quadrant, and mixing them up is the single most common ASTC mistake. $\times$ (cos is not positive in Quadrant II)
2.4 Reference Angles and Exact Values
Worked Example 2: Find $\cos(210°)$ and $\sin(210°)$ using a reference angle.
$210°$ is in Quadrant III ($180°$ to $270°$).
Reference angle (angle to the nearest $x$-axis): $210° - 180° = 30°$
Quadrant III signs: sin negative, cos negative (from the ASTC table).
Magnitudes match the $30°$ reference angle: $\sin30°=1/2$, $\cos30°=\sqrt{3}/2$
\[\cos(210°) = -\frac{\sqrt{3}}{2} \qquad \sin(210°) = -\frac{1}{2}\]Worked Example 3: Find $\tan(5\pi/3)$.
$5\pi/3$ radians $= 300°$ (since $5\pi/3 \times 180/\pi = 900/3 = 300°$)
Quadrant IV ($270°$ to $360°$).
Reference angle: $360° - 300° = 60°$
Quadrant IV: tan is negative (from ASTC table). $\tan(60°) = \sqrt{3}$
\[\tan(5\pi/3) = -\sqrt{3}\]2.5 Standard Values Table
| $\theta$ | $0$ | $\pi/6$ | $\pi/4$ | $\pi/3$ | $\pi/2$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $1/2$ | $\sqrt{2}/2$ | $\sqrt{3}/2$ | $1$ |
| $\cos\theta$ | $1$ | $\sqrt{3}/2$ | $\sqrt{2}/2$ | $1/2$ | $0$ |
| $\tan\theta$ | $0$ | $\sqrt{3}/3$ | $1$ | $\sqrt{3}$ | undefined |
3. Graphs of Trig Functions
| Function | Period | Domain | Range | Key features |
|---|---|---|---|---|
| $\sin x$ | $2\pi$ | $\mathbb{R}$ | $[-1, 1]$ | Passes through $(0,0)$, odd function |
| $\cos x$ | $2\pi$ | $\mathbb{R}$ | $[-1, 1]$ | Passes through $(0,1)$, even function |
| $\tan x$ | $\pi$ | $x \neq \pi/2 + k\pi$ | $\mathbb{R}$ | Vertical asymptotes where $\cos x = 0$ |
Amplitude/period/shift form: $y = A\sin(Bx - C) + D$ has amplitude $\lvert A \rvert$, period $2\pi/\lvert B \rvert$, phase shift $C/B$, and vertical shift $D$.
Worked Example 4: State the amplitude, period, phase shift, and vertical shift of $y = 3\sin(2x - \pi/2) + 1$.
\(A = 3 \to \text{amplitude } 3 \qquad B = 2 \to \text{period} = 2\pi/2 = \pi\) \(C = \pi/2 \to \text{phase shift} = C/B = (\pi/2)/2 = \pi/4 \text{ (shift right)} \qquad D = 1 \to \text{vertical shift up 1}\)
4. Fundamental Identities
4.1 Pythagorean Identities
From $x^2 + y^2 = 1$ (the unit circle equation) with $x = \cos\theta$, $y = \sin\theta$:
\[\sin^2\theta + \cos^2\theta = 1\]Dividing through by $\cos^2\theta$ and by $\sin^2\theta$ respectively gives two more:
\[\tan^2\theta + 1 = \sec^2\theta \qquad 1 + \cot^2\theta = \csc^2\theta\]Worked Example 5: Given $\sin\theta = 3/5$ with $\theta$ in Quadrant II, find $\cos\theta$ and $\tan\theta$.
\[\sin^2\theta + \cos^2\theta = 1 \implies (3/5)^2 + \cos^2\theta = 1 \implies \frac{9}{25} + \cos^2\theta = 1 \implies \cos^2\theta = \frac{16}{25} \implies \cos\theta = \pm\frac{4}{5}\]Quadrant II: cos is negative (ASTC table) $\to \cos\theta = -4/5$
\[\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}\]Example 3 (Pythagorean identity holds for any θ): Verify $\sin^2\theta + \cos^2\theta = 1$ at $\theta = \pi/4$.
$\sin(\pi/4) = \sqrt{2}/2$, $\cos(\pi/4) = \sqrt{2}/2$. $\sin^2(\pi/4) + \cos^2(\pi/4) = (\sqrt{2}/2)^2 + (\sqrt{2}/2)^2 = 1/2 + 1/2 = 1$. $\checkmark$ Matches — and this will check out at EVERY $\theta$, since it comes directly from the unit-circle equation $x^2+y^2=1$, not from a coincidence at this particular angle.
Non-Example 3 (A look-alike formula that is NOT an identity): Is $\sin\theta + \cos\theta = 1$ true for every $\theta$?
Test $\theta = \pi/4$: $\sin(\pi/4) + \cos(\pi/4) = \sqrt{2}/2 + \sqrt{2}/2 = \sqrt{2} \approx 1.414$. Is $\sqrt{2} = 1$? No. $\times$ It happens to hold at $\theta = 0$ ($\sin 0+\cos 0 = 0+1 = 1$) and at $\theta = \pi/2$ ($1+0=1$), but FAILS at $\pi/4$ — one counterexample is enough to disqualify it as an identity. “$\sin\theta + \cos\theta = 1$” is only an EQUATION, true for a couple of specific $\theta$ values, never a general identity like $\sin^2\theta+\cos^2\theta=1$.
4.2 Sum and Difference Formulas
\(\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B\) \(\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B \quad \text{(note the sign flips)}\) \(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}\)
Worked Example 6: Find the exact value of $\sin(75°)$ using the sum formula with $75° = 45° + 30°$.
\(\sin(75°) = \sin(45° + 30°) = \sin45°\cos30° + \cos45°\sin30°\) \(= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)
4.3 Double Angle Formulas
\(\sin(2A) = 2\sin A\cos A\) \(\cos(2A) = \cos^2A - \sin^2A = 2\cos^2A - 1 = 1 - 2\sin^2A\) \(\tan(2A) = \frac{2\tan A}{1 - \tan^2A}\)
Worked Example 7: Given $\sin A = 5/13$ with $A$ in Quadrant I, find $\sin(2A)$ and $\cos(2A)$.
Quadrant I: everything positive.
$\cos A$: $\sin^2A + \cos^2A = 1 \implies \cos^2A = 1 - 25/169 = 144/169 \implies \cos A = 12/13$
\(\sin(2A) = 2\sin A\cos A = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169}\) \(\cos(2A) = 1 - 2\sin^2A = 1 - 2\left(\frac{25}{169}\right) = 1 - \frac{50}{169} = \frac{119}{169}\)
4.4 Identity vs. Equation
An identity (like $\sin^2\theta + \cos^2\theta = 1$) is true for every value of $\theta$ in the domain — it’s proven by algebraic manipulation, never by “solving for $\theta$.” An equation (like $\sin\theta = 1/2$) is only true for specific values of $\theta$, found by solving.
Worked Example 8: Verify (prove) the identity $(1 - \cos^2\theta)/\sin\theta = \sin\theta$.
\(\text{LHS} = \frac{1 - \cos^2\theta}{\sin\theta} = \frac{\sin^2\theta}{\sin\theta} \quad \text{[Pythagorean identity: } 1 - \cos^2\theta = \sin^2\theta\text{]}\) \(= \sin\theta \quad \text{[cancel one factor of } \sin\theta, \ \theta \neq k\pi\text{]} = \text{RHS} \ \checkmark\)
This holds for every $\theta$ where $\sin\theta \neq 0$ — it’s an identity, not something to “solve for $\theta$.”
Example 4 (Identity — true at every θ tested): Check $\sin(2\theta) = 2\sin\theta\cos\theta$ at $\theta = \pi/6$ and at $\theta = \pi/3$.
$\theta = \pi/6$: LHS $= \sin(\pi/3) = \sqrt{3}/2$. RHS $= 2\sin(\pi/6)\cos(\pi/6) = 2(1/2)(\sqrt{3}/2) = \sqrt{3}/2$. LHS $=$ RHS $\checkmark$
$\theta = \pi/3$: LHS $= \sin(2\pi/3) = \sqrt{3}/2$. RHS $= 2\sin(\pi/3)\cos(\pi/3) = 2(\sqrt{3}/2)(1/2) = \sqrt{3}/2$. LHS $=$ RHS $\checkmark$
Holds at both (different) angles tested — consistent with it being an identity, true for every $\theta$ (it’s the double angle formula from Section 4.3).
Non-Example 4 (Equation — true only at specific θ): Check $\sin\theta = 1/2$ at $\theta = \pi/6$ and at $\theta = \pi/3$.
$\theta = \pi/6$: $\sin(\pi/6) = 1/2$. Is $1/2 = 1/2$? Yes. $\checkmark$ holds here
$\theta = \pi/3$: $\sin(\pi/3) = \sqrt{3}/2 \approx 0.866$. Is $\sqrt{3}/2 = 1/2$? No. $\times$ fails here
It holds at one angle and fails at another — this is the signature of an EQUATION, not an identity: it constrains $\theta$ to a specific set of solutions ($\theta = \pi/6, 5\pi/6, \ldots$) rather than holding universally.
5. Solving Trigonometric Equations
General approach: isolate the trig function, find all solutions in one period using reference angles/known values and the ASTC sign pattern, then add the period times any integer $k$ to capture every solution (unless a restricted domain is given).
Worked Example 9: Solve $2\sin\theta - 1 = 0$ for $\theta \in [0, 2\pi)$.
\[2\sin\theta = 1 \implies \sin\theta = \frac{1}{2}\]Reference angle: $\pi/6$ (since $\sin(\pi/6) = 1/2$). sin is positive in Quadrants I and II (ASTC table).
Quadrant I: $\theta = \pi/6$. Quadrant II: $\theta = \pi - \pi/6 = 5\pi/6$
Solutions in $[0, 2\pi)$: $\theta = \pi/6, 5\pi/6$
Worked Example 10: Solve $2\cos^2\theta - 1 = 0$ for $\theta \in [0, 2\pi)$.
\[\cos^2\theta = \frac{1}{2} \implies \cos\theta = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}\]Reference angle: $\pi/4$
$\cos\theta = \sqrt{2}/2$: Quadrants I, IV (cos positive) $\to \theta = \pi/4, 7\pi/4$
$\cos\theta = -\sqrt{2}/2$: Quadrants II, III (cos negative) $\to \theta = 3\pi/4, 5\pi/4$
Solutions: $\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$
Worked Example 11: Solve $2\sin^2\theta + \sin\theta - 1 = 0$ for $\theta \in [0, 2\pi)$.
Let $u = \sin\theta$. Then $2u^2 + u - 1 = 0$. Factor: $(2u - 1)(u + 1) = 0 \implies u = 1/2$ or $u = -1$
Case $u = 1/2$: $\sin\theta = 1/2 \to \theta = \pi/6, 5\pi/6$ (as in Example 9)
Case $u = -1$: $\sin\theta = -1 \to \theta = 3\pi/2$ (only one solution: minimum of sine)
Solutions: $\theta = \pi/6, 5\pi/6, 3\pi/2$
Example 5 (Correct method — no solutions lost): Solve $\sin(2\theta) = \sin\theta$ for $\theta \in [0, 2\pi)$.
Move everything to one side FIRST — do not divide by $\sin\theta$: \(\sin(2\theta) - \sin\theta = 0 \implies 2\sin\theta\cos\theta - \sin\theta = 0 \quad \text{[double angle formula]} \implies \sin\theta(2\cos\theta - 1) = 0 \quad \text{[factor out } \sin\theta\text{]}\)
Case $\sin\theta = 0$: $\theta = 0, \pi$. Case $2\cos\theta - 1 = 0$, i.e. $\cos\theta = 1/2$: $\theta = \pi/3, 5\pi/3$
Solutions: $\theta = 0, \pi/3, \pi, 5\pi/3$ (four solutions, including the $\sin\theta = 0$ branch)
Non-Example 5 (Dividing by sin θ silently loses solutions): Solve the same equation $\sin(2\theta) = \sin\theta$ by dividing both sides by $\sin\theta$ instead.
\[\sin(2\theta) = \sin\theta \implies 2\sin\theta\cos\theta = \sin\theta\]Divide both sides by $\sin\theta$: $2\cos\theta = 1 \implies \cos\theta = 1/2 \implies \theta = \pi/3, 5\pi/3$
Compare with the correct solution set $\{0, \pi/3, \pi, 5\pi/3\}$ above: $\theta = 0$ and $\theta = \pi$ are MISSING. $\times$ Why: dividing by $\sin\theta$ silently assumes $\sin\theta \neq 0$, which throws away exactly the solutions where $\sin\theta = 0$. Dividing an equation by an expression that could be zero is never safe — factor instead, so every case (including “the factor equals 0”) stays visible.
6. Inverse Trigonometric Functions
6.1 Why Restriction Is Needed
sin, cos, tan are not injective on $\mathbb{R}$ (Module 0b, Section 2.1) — infinitely many angles share the same sine, for instance. To define an inverse (Module 0b, Section 4.1), each function is restricted to an interval where it is bijective onto its range.
| Function | Restricted domain | Range of inverse |
|---|---|---|
| $\sin^{-1}x$ (arcsin) | $[-\pi/2, \pi/2]$ | $[-1, 1] \to [-\pi/2, \pi/2]$ |
| $\cos^{-1}x$ (arccos) | $[0, \pi]$ | $[-1, 1] \to [0, \pi]$ |
| $\tan^{-1}x$ (arctan) | $(-\pi/2, \pi/2)$ | $\mathbb{R} \to (-\pi/2, \pi/2)$ |
Worked Example 12: Evaluate $\sin^{-1}(1/2)$, $\cos^{-1}(-1/2)$, $\tan^{-1}(1)$.
$\sin^{-1}(1/2)$: need $\theta \in [-\pi/2, \pi/2]$ with $\sin\theta = 1/2$. $\theta = \pi/6$. $\checkmark$ (in range)
$\cos^{-1}(-1/2)$: need $\theta \in [0, \pi]$ with $\cos\theta = -1/2$.
Reference angle $\pi/3$; cos negative $\to$ Quadrant II angle $\to \theta = \pi - \pi/3 = 2\pi/3$. $\checkmark$ (in $[0,\pi]$)
$\tan^{-1}(1)$: need $\theta \in (-\pi/2, \pi/2)$ with $\tan\theta = 1$. $\theta = \pi/4$. $\checkmark$
Example 6 (Correct inverse trig value — inside the restricted range): Evaluate $\sin^{-1}(1/2)$.
Need $\theta \in [-\pi/2, \pi/2]$ (the restricted domain for arcsin) with $\sin\theta = 1/2$. $\theta = \pi/6$ satisfies $\sin(\pi/6) = 1/2$, and $\pi/6 \in [-\pi/2, \pi/2]$. $\checkmark$ $\sin^{-1}(1/2) = \pi/6$ — this IS the answer, because it’s both correct and inside the allowed range.
Non-Example 6 (Correct sine value, but outside the restricted range — NOT the answer): Is $\theta = 5\pi/6$ a valid value for $\sin^{-1}(1/2)$?
Check $\sin(5\pi/6)$: $5\pi/6$ is in Quadrant II, reference angle $\pi/6$, sin positive there. $\sin(5\pi/6) = 1/2$. The sine value is correct! But is $5\pi/6 \in [-\pi/2, \pi/2]$ (the restricted range for arcsin)? $\pi/2 \approx 1.57$, $5\pi/6 \approx 2.62$. Is $2.62 \leq 1.57$? No. $\times$ Even though $\sin(5\pi/6) = 1/2$ is true, $5\pi/6$ is OUTSIDE arcsin’s allowed range, so it is NOT the value of $\sin^{-1}(1/2)$ — the inverse function must return the ONE angle inside the restricted domain, and that is $\pi/6$, not $5\pi/6$.
Key habit: unlike Worked Examples 9–11 (which wanted all solutions in $[0, 2\pi)$), an inverse trig function returns exactly one value — the one lying in its restricted range. Don’t list multiple answers for $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$.
7. Summary
| Concept | Key Idea | Worked Example |
|---|---|---|
| Radians ↔ degrees | Multiply by $\pi/180$ or $180/\pi$ | $150° = 5\pi/6$ |
| Unit circle definition | $\cos\theta = x$, $\sin\theta = y$ on unit circle | Works for all real $\theta$ |
| ASTC | Sign of sin/cos/tan by quadrant | $\cos(210°) = -\sqrt{3}/2$ |
| Reference angle | Acute angle to $x$-axis, use with ASTC sign | $\tan(5\pi/3) = -\sqrt{3}$ |
| Pythagorean identity | $\sin^2\theta+\cos^2\theta=1$, and its two variants | Find $\cos\theta, \tan\theta$ from $\sin\theta$ |
| Sum/difference formulas | $\sin(A\pm B)$, $\cos(A\pm B)$ | $\sin75° = (\sqrt{6}+\sqrt{2})/4$ |
| Double angle formulas | $\sin2A=2\sin A\cos A$, etc. | $\sin2A, \cos2A$ from $\sin A=5/13$ |
| Identity vs. equation | Identity: true always. Equation: solve for $\theta$ | $(1-\cos^2\theta)/\sin\theta = \sin\theta$, proven |
| Solving trig equations | Reference angle + ASTC + period | $2\sin^2\theta+\sin\theta-1=0 \implies 3$ solutions |
| Inverse trig functions | Restricted domain makes the function invertible; ONE answer only | $\sin^{-1}(1/2) = \pi/6$ |
8. Practice Problems
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Convert $315°$ to radians and $7\pi/4$ radians to degrees, and confirm they match.
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Find the exact value of $\sin(135°)$ and $\cos(135°)$ using a reference angle.
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Find the exact value of $\tan(-\pi/3)$ using a reference angle and the sign for its quadrant.
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Given $\cos\theta = -5/13$ with $\theta$ in Quadrant III, find $\sin\theta$ and $\tan\theta$.
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Given $\tan\theta = 4/3$ with $\theta$ in Quadrant I, find $\sin\theta, \cos\theta$ using the Pythagorean identity $\tan^2\theta + 1 = \sec^2\theta$ (find $\sec\theta$ first, then $\cos\theta$, then $\sin\theta$).
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Find the exact value of $\cos(15°)$ using the difference formula with $15° = 45° - 30°$.
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Find the exact value of $\tan(105°)$ using the sum formula with $105° = 60° + 45°$.
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Given $\cos A = 3/5$ with $A$ in Quadrant I, find $\sin(2A)$ and $\cos(2A)$.
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Prove the identity: $\sec^2\theta - 1 = \tan^2\theta$ (starting from the Pythagorean identity).
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Prove the identity: $\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} = \dfrac{2}{\sin\theta}$. (Hint: combine over a common denominator, then use $\sin^2\theta + \cos^2\theta = 1$ in the numerator.)
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Solve $2\cos\theta + 1 = 0$ for $\theta \in [0, 2\pi)$.
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Solve $\tan\theta = \sqrt{3}$ for $\theta \in [0, 2\pi)$.
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Solve $2\sin^2\theta - 3\sin\theta + 1 = 0$ for $\theta \in [0, 2\pi)$. (Hint: let $u = \sin\theta$, factor the quadratic in $u$ first.)
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Solve $\cos(2\theta) = \cos\theta$ for $\theta \in [0, 2\pi)$. (Hint: rewrite $\cos(2\theta)$ using a double angle formula in terms of $\cos\theta$ alone, then factor.)
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Evaluate $\sin^{-1}(-\sqrt{3}/2)$, $\cos^{-1}(0)$, $\tan^{-1}(-1)$, giving your answer in the correct restricted range for each.
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State the amplitude, period, phase shift, and vertical shift of $y = -2\cos(3x + \pi) - 4$.
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(Harder) Solve $\sin(2\theta) = \sin\theta$ for $\theta \in [0, 2\pi)$. (Hint: move everything to one side and factor — do not divide both sides by $\sin\theta$, since that can silently discard solutions where $\sin\theta = 0$.)
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(Harder) Evaluate $\cos(\sin^{-1}(3/5))$ without a calculator, by drawing a right triangle for the inner inverse-sine angle and reading off the cosine.