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.)

InverseDomain (argument)Range (principal value)Quadrants allowedSign 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)))

  1. Reduce the raw angle to an equivalent between 0 and 2\pi (or between -\pi and \pi for odd functions).
  2. Identify its quadrant.
  3. 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.
  4. 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
  1. \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).
  2. \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
  1. Replace f(x) with y.
  2. Switch roles of x and y.
  3. Solve the new equation for y (isolate y, keep it positive if possible).
  4. 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.