Comprehensive Trigonometry Review – Inverses, Identities & Complementary Angles

Quiz Logistics & Expectations

• Quizzes are versioned; always state the question number and version when asking for help.
• Show every algebraic step; calculator-only answers earn no credit.
• For trigonometric simplifications, write down the identity you are using (e.g. $1+\tan^2\theta=\sec^2\theta$) before substituting.
• On the upcoming test you may skip exactly one problem. Instructor recommends skipping the longest problem to maximize time.
• Participation matters: active students receive small grade boosts (e.g. 68→70). Silent attendance does not.

Inverse Trig Functions & Domain-Range Swapping

General Workflow

1. Swap variables $x \leftrightarrow y$.
2. Isolate $y$ algebraically (add/subtract constants, divide coefficients, apply inverse trig).
3. Identify the inner expression of the inverse trig; its natural domain controls the new domain.
4. Solve the resulting compound inequality for $x$.

Example 1 (y=4-\cos(3x-12))

1. Swap: $x = 4-\cos(3y-12)$.
2. Isolate:
   • $x-4 = -\cos(3y-12)$
   • $-(x-4) = \cos(3y-12)$
   • $3y-12 = \cos^{-1}!\bigl(-(x-4)\bigr)$
   • $3y = \cos^{-1}!\bigl(-(x-4)\bigr)+12$
   • $y = -\tfrac13\cos^{-1}(x-4)+4$
3. Since $\cos^{-1}$ requires $-1\le (x-4)\le1$:
   • $-1\le x-4 \le 1\;\;\Rightarrow\;\;3\le x\le5$ ← domain of the inverse.

Example 2 (y=7+5\sin(2x+9))

1. Swap: $x = 7+5\sin(2y+9)$
2. Isolate:
   $\sin(2y+9)=\dfrac{x-7}{5}$
   $2y+9 = \sin^{-1}!\left(\dfrac{x-7}{5}\right)$
   $y = -\tfrac92+\tfrac12\sin^{-1}!\left(\dfrac{x-7}{5}\right)$
3. $\sin^{-1}$ demands $-1\le\dfrac{x-7}{5}\le1$ ⇨ $2\le x\le12$.

Key Reminders

• Natural domains:
  • $\sin^{-1},\;\cos^{-1}:\;[-1,1]$
  • $\tan^{-1}:\;(-\infty,\infty)$ (unchanged when taking inverses).
• Do not shortcut by guessing range→domain; coefficients in front of trig terms can invalidate the guess.

Fundamental Trig Identities (used repeatedly)

• Pythagorean set
  $\sin^2\theta+\cos^2\theta=1$
  $1+\tan^2\theta=\sec^2\theta$
  $1+\cot^2\theta=\csc^2\theta$
• Cofunction (complementary-angle) set
  $\sin(90^\circ-\theta)=\cos\theta$\;; $\cos(90^\circ-\theta)=\sin\theta$
  $\tan(90^\circ-\theta)=\cot\theta$\;; $\cot(90^\circ-\theta)=\tan\theta$
  $\sec(90^\circ-\theta)=\csc\theta$\;; $\csc(90^\circ-\theta)=\sec\theta$
• Angle-sum/difference (needed for calculator-free exact values)
  $\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
  $\cos(A-B)=\cos A\cos B+\sin A\sin B$ (sign flips for $+$)

Complementary-Angle Theorem in Practice

Given two acute angles $\alpha,\beta$ in a right triangle: $\alpha+\beta=90^\circ$.
Therefore any cofunction pair shares a value:

• $\sin 37^\circ = \cos 53^\circ$,
• $\tan^2 23^\circ = \cot^2 67^\circ$, etc.

Fast Evaluations

• $\sin^2 18^\circ+\cos^2 72^\circ = 1$ (same angle after complement)
• $\csc^2 10^\circ-\cot^2 80^\circ = 1$
• Mixed powers example
  $\sin^4 70^\circ-\cot^2 3^\circ+\sec^2 77^\circ$
  • First two terms form $1$
  • Remaining pair by identity ⇒ $1$
  • Combined sign ⇒ final $1$.

Radian Variant

• Complement of $3\pi/8$ is $\pi/2-3\pi/8=\pi/8$.
• Example simplification
  $\sin^2!\frac{3\pi}{8}+\tan^2!\frac{3\pi}{8}+\cos^2!\frac{\pi}{8}-\sec^2!\frac{3\pi}{8}$
  → First and third form $1$; last pair gives $-1$; result $0$.

Exact-Value Problems Using Reference Triangles

Setup Guidelines

1. Interpret interval to place the angle in the correct quadrant.
2. Draw right triangle; assign signs (+/–) to legs.
3. Use given trig ratio (e.g. $\tan\alpha=-36/77$) to label opposite/adjacent.
4. Apply Pythagorean theorem to find hypotenuse.
5. Extract all six trig values for that angle.

Example

Given $\tan\alpha=-\tfrac{36}{77}$ with $\alpha\in(\pi,3\pi/2)$ (Q III) and $\cos\beta=\tfrac{\sqrt6}{3}$ with $\beta\in(0,\tfrac\pi2)$ (Q I), find $\cos(\alpha-\beta)$.

• Triangle for $\alpha$: opposite $36$, adjacent $-77$ ⇒ hyp $85$.
  $\cos\alpha=-77/85,\;\sin\alpha=36/85$
• Triangle for $\beta$: adjacent $\sqrt6$, hyp $3$ ⇒ opposite $\sqrt3$.
  $\cos\beta=\sqrt6/3,\;\sin\beta=\sqrt3/3$
• Apply formula:
  $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$
  $=(-\frac{77}{85})(\frac{\sqrt6}{3})+(\frac{36}{85})(\frac{\sqrt3}{3})$
  $=\frac{-77\sqrt6+36\sqrt3}{255}$.

Identity Simplification Walk-Through (Instructor’s Board Example)

Expression: $\sec^2\theta-\tan^2\theta\;\;\bigl(=1\bigr)$

1. Recognize direct Pythagorean identity → equals $1$ immediately.
2. Alternative: rewrite sec & tan in $\sin,\cos$, find common denominator, cancel.

Takeaway: multiple valid paths; credit awarded if algebra is sound.

Instructor Tips & Warnings

• Never submit a raw calculator decimal (e.g. entering $\tan(342^\circ)$ and writing $-0.532\dots$). Write angle-sum identity first, then exact radical.
• Typical red flag: student writes $173$ where only $\sqrt3/2$ is correct.
• Time management: familiar topics → fast, unfamiliar → stall. Master the entire review sheet.
• Allowed resources in online setting: notes, book, videos, friends; nevertheless, without practice time will expire.

Miscellaneous Reminders

• Degree ↔ Radian: $90^\circ = \pi/2$.
• Always keep $2k\pi$ terms when solving trig equations; you can clear denominators later so they share the same base.
• When multiplying an entire equation by a common denominator, distribute to every term including $2k\pi$.

Ethical / Philosophical Nuggets

• “Show the process or receive no credit” – mathematics values transparency.
• Quizzes emulate real-life problem solving where intermediate reasoning is as important as answers.
• Instructor’s closing note: personal encouragement and faith reference (invitation to “let Jesus live in your heart”).