Inverse Trigonometric Functions & Algebraic Inverses – Lecture Notes

Announcements

Test #1
- Instructor still grading; answer key to be posted Monday afternoon.
- Tuesday’s class will be devoted to going over the key and discussing grade-change requests.
Current lecture covers Section 3.1 (pp. 195–197 in “old” text): inverse trig functions & algebraic inverses.
Practice worksheet on composite functions already in D2L; last three items use “ugly numbers” (non-standard denominators).

Quick Reference – Principal‐Value Ranges & Allowed Quadrants

(Always check whether the quadrant produced by the raw angle is inside this range; if not, reflect/translate.)

Inverse	Domain (argument)	Range (principal value)	Quadrants allowed	Sign of original trig fn in that range
$\sin^{-1}(x)$	$[-1,1]$	$[-\tfrac{\pi}{2},\;\tfrac{\pi}{2}]$	I & IV	+ in I, – in IV
$\cos^{-1}(x)$	$[-1,1]$	$[0,\;\pi]$	I & II	+ in I, – in II
$\tan^{-1}(x)$	$(-\infty,\infty)$	$(-\tfrac{\pi}{2},\;\tfrac{\pi}{2})$	I & IV	+ in I, – in IV
$\cot^{-1}(x)$	$(-\infty,\infty)$	$(0,\;\pi)$	I & II	+ in I, – in II
$\sec^{-1}(x)$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus{\tfrac{\pi}{2}}$	I (positive), II (negative)
$\csc^{-1}(x)$	$(-\infty,-1]\cup[1,\infty)$	$[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\setminus{0}$	IV (negative), I (positive)

Strategy for “Ugly‐Denominator” Composite Problems (e.g.

$\cos^{-1}(\cos(8\pi/7))$ )

Reduce the raw angle to an equivalent between $0$ and $2\pi$ (or between $-\pi$ and $\pi$ for odd functions).
Identify its quadrant.
Check whether that quadrant belongs to the inverse function’s principal range.
- If yes, you may still have to switch to the negative representation (quadrant IV) for $\tan^{-1}$ or $\sin^{-1}$ to keep within the OPEN interval at the bottom boundary.
- If no, measure the distance to the nearest boundary of the allowed range and “reflect” the point across that boundary.
Express the final answer in EXACT radians; keep the same denominator the problem started with.

Term used in class: angles with unfamiliar denominators (7, 11, 17, 26, …) are “ugly numbers.”

Detailed Worked Examples from Class

1. $\cos^{-1}\bigl(\cos(8\pi/7)\bigr)$

$8\pi/7$ lies slightly past $\pi$ (because $7\pi/7=\pi$ ) ⇒ Quadrant III.
Principal range of $\cos^{-1}$ is Quadrants I & II.
Distance method: $8\pi/7-7\pi/7=\pi/7$ . Reflect left of $\pi$ by same $\pi/7$ ⇒ $\pi-\pi/7=6\pi/7.$
Result: $\boxed{\;6\pi/7\;}.$

2. $\tan^{-1}\bigl(\tan(31\pi/17)\bigr)$

$17\pi/17=\pi$ , $34\pi/17=2\pi$ . Angle sits in Quadrant IV (between $\tfrac{25.5\pi}{17}$ and $2\pi$ ).
Range of $\tan^{-1}$ is $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$ ⇒ Quadrant IV but must be expressed as a negative measure.
Distance to $2\pi$ : $34\pi/17-31\pi/17=3\pi/17$ ⇒ answer $-3\pi/17.$
Alternative positive coterminal $=34\pi/17-3\pi/17=31\pi/17$ , but principal value is $\boxed{-3\pi/17}.$

3. $\cot^{-1}\bigl(\cot(-2\pi/11)\bigr)$

Clockwise ⇒ Quadrant IV.
$\cot^{-1}$ range Quadrants I & II.
Need equivalent in Quadrant II with same absolute value of cotangent.
Symmetric reflection through x-axis or add $\pi$ : choices given in class:
- $\boxed{9\pi/11}\;(=\pi-2\pi/11)$
- or its negative coterminal $\boxed{-13\pi/11}$ (both describe the same location).

4. $\sin^{-1}\bigl(\sin(-17\pi/26)\bigr)$

$-17\pi/26$ ≈ −117.7° → Quadrant III going clockwise.
$\sin^{-1}$ range: Quadrants IV & I.
Reflect into Quadrant IV keeping distance 4π/26: answer $-9\pi/26$ (coterminal positive $43\pi/26$ ).
Principal value within $[-\pi/2,\pi/2]$ is $\boxed{-9\pi/26}.$

Practice Questions Discussed

$\tan^{-1}\bigl(\tan(7\pi/10)\bigr)$
- Raw angle Quadrant II; subtract $\pi$ → $-3\pi/10$ . Class wrote equivalent $17\pi/10$ . Principal answer: $\boxed{-3\pi/10}\;(=17\pi/10).$
$\sec^{-1}\bigl(\sec(-4\pi/13)\bigr)$
- Quadrant IV (secant positive). Needed Quadrant I → $\boxed{4\pi/13}.$

Algebraic Inverses of Functions (pp. 197 ff.)

Standard 4-Step Procedure

Replace $f(x)$ with $y$ .
Switch roles of $x$ and $y$ .
Solve the new equation for $y$ (isolate $y$ , keep it positive if possible).
Rename $y$ as $f^{-1}(x)$ and state the domain.

Example A

Given $f(x)=5\cos x+3$ .

Steps ⇒ $f^{-1}(x)=\cos^{-1}\Bigl(\dfrac{x-3}{5}\Bigr).$
Domain: start from $-1\le\dfrac{x-3}{5}\le1$ ⇒ $-2\le x\le8.$

Example B

$f(x)=73\,\tan(15x-7)+13$

$f^{-1}(x)=\dfrac{1}{15}\tan^{-1}\Bigl(\dfrac{x-13}{73}\Bigr)+\dfrac{7}{15}.$
Range of $\tan^{-1}$ is unbounded ⇒ domain remains $(-\infty,\infty).$

Example C

$f(x)=40+\sin(3x+8)$

After algebra: $f^{-1}(x)=\dfrac{1}{3}\sin^{-1}(x-40)-\dfrac{8}{3}.$
Domain from $-1\le x-40\le1$ ⇒ $39\le x\le41.$

Recurring Algebra Tricks & Pitfalls

Never distribute constants into an inverse‐trig parenthesis; keep the inside expression intact until final isolation.
When moving a trig function from denominator to numerator, change it to its inverse (e.g.
$\frac{1}{\cos}\;\rightarrow\;\cos^{-1}$ ).
For $\tan^{-1}$ and $\cot^{-1}$ principal ranges are open at the endpoints; keep results strictly inside.
After finding an inverse, always impose the domain by turning the range restriction of the inverse trig function into a double inequality and solving for $x$ .

Pedagogical / Administrative Notes

Only ~5 of ~50 students were present live; instructor expressed concern.
Emphasis on working through the D2L practice set before the upcoming quiz and test (especially the final three “ugly number” problems).
Next class (Tuesday) starts with any questions on the posted answer key for Test #1.

Key Takeaways

“Distance & reflect” is the go-to method for composite inverse-trig problems with uncommon denominators.
Memorize principal-value ranges and quadrant sign rules; they dictate every conversion.
For algebraic inverses: swap variables, isolate $y$ , apply inverse operations systematically, then enforce domains via the inverse’s range.
Always express final answers in exact radian form, keeping the original denominator when possible.