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5.5 Inverse Trigonometric Functions and Their Graphs

Focus on principal inverse sine and inverse sine domain.

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Ideas in this lesson

Explore the lesson brief and the worked example for Idea 11. Create a free account to continue with the remaining ideas and practice.

01Restricted inverse definitionsWorked steps · five questions02Exact and numerical inverse evaluationWorked steps · five questions03Cancellation with domainsWorked steps · five questions04Mixed trig-inverse compositionsWorked steps · five questions05Inverse identity conjectures and proofWorked steps · five questions06Compare compositions globallyWorked steps · five questions07Inverse reciprocal branchesWorked steps · five questions
Lesson brief and rules

5.5 Inverse Trigonometric Functions and Their Graphs

arcsin⁡\arcsin returns a principal angle in [−π2,π2][-\frac{\pi }{2},\frac{\pi }{2}].

Worked example 11
01 / Explain why the full sin⁡\sin function has no inverse function, and why restricting its input to [−π2,π2][-\dfrac\pi2,\dfrac\pi2] works. Read the graph of the restricted function (( blue )), inverse (( pink )), and reflection line (( green )). State the inverse domain/range, included endpoints or horizontal asymptotes, and evaluate arcsin⁡(13)\arcsin(\dfrac{1}{3}) to five decimals. Distinguish inverse notation from a reciprocal.
Restricted function, inverse and reflection lineHorizontal and vertical coordinates use numeric ticks rounded to three decimals. Curves are clipped to the stated window.−4−2.4−0.80.82.44−4−2.4−0.80.82.44
Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
02 / The unrestricted function repeats outputs, so it fails the horizontal line test. On [−π2,π2][-\dfrac\pi2,\dfrac\pi2] it is strictly increasing and takes every output of its original range once.
03 / Reflection swaps input and output: arcsin⁡(u)=w⟺sin⁡(w)=u\arcsin(u)=w\quad\Longleftrightarrow\quad\sin(w)=u with w∈[−π2,π2]w\in[-\dfrac\pi2,\dfrac\pi2]. Hence the inverse domain is [−1,1][-1,1] and its range is [−π2,π2][-\dfrac\pi2,\dfrac\pi2].
04 / The two endpoints are included, and there are no horizontal asymptotes.
05 / arcsin⁡(13)≈0.33984\arcsin(\dfrac{1}{3})\approx 0.33984 in radians. sin⁡−1(u)\sin^{-1}(u) denotes this inverse here, not 1sin⁡u\dfrac1{\sin u}.
Worked example 2 · Restricted inverse definitions

Explain why the full cos⁡\cos function has no inverse function, and why restricting its input to [0,π][0,\pi ] works. Read the graph of the restricted function (( blue )), inverse (( pink )), and reflection line (( green )). State the inverse domain/range, included endpoints or horizontal asymptotes, and evaluate arccos⁡(12)\arccos(\dfrac{1}{2}) to five decimals. Distinguish inverse notation from a reciprocal.

Restricted function, inverse and reflection lineHorizontal and vertical coordinates use numeric ticks rounded to three decimals. Curves are clipped to the stated window.−4−2.4−0.80.82.44−4−2.4−0.80.82.44
Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

The unrestricted function repeats outputs, so it fails the horizontal line test. On [0,π][0,\pi ] it is strictly decreasing and takes every output of its original range once.
Reflection swaps input and output: arccos⁡(u)=w⟺cos⁡(w)=u\arccos(u)=w\quad\Longleftrightarrow\quad\cos(w)=u with w∈[0,π]w\in[0,\pi ]. Hence the inverse domain is [−1,1][-1,1] and its range is [0,π][0,\pi ].
The two endpoints are included, and there are no horizontal asymptotes.
arccos⁡(12)≈1.04720\arccos(\dfrac{1}{2})\approx 1.04720 in radians. cos⁡−1(u)\cos^{-1}(u) denotes this inverse here, not 1cos⁡u\dfrac1{\cos u}.
Written-work review:

Read the graphs

Restricted inverse definitions

Explain why the full sin⁡\sin function has no inverse function, and why restricting its input to [−π2,π2][-\dfrac\pi2,\dfrac\pi2] works. Read the graph of the restricted function (( blue )), inverse (( pink )), and reflection line (( green )). State the inverse domain/range, included endpoints or horizontal asymptotes, and evaluate arcsin⁡(13)\arcsin(\dfrac{1}{3}) to five decimals. Distinguish inverse notation from a reciprocal.
Restricted function, inverse and reflection lineHorizontal and vertical coordinates use numeric ticks rounded to three decimals. Curves are clipped to the stated window.−4−2.4−0.80.82.44−4−2.4−0.80.82.44
Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

The unrestricted function repeats outputs, so it fails the horizontal line test. On [−π2,π2][-\dfrac\pi2,\dfrac\pi2] it is strictly increasing and takes every output of its original range once.
Reflection swaps input and output: arcsin⁡(u)=w⟺sin⁡(w)=u\arcsin(u)=w\quad\Longleftrightarrow\quad\sin(w)=u with w∈[−π2,π2]w\in[-\dfrac\pi2,\dfrac\pi2]. Hence the inverse domain is [−1,1][-1,1] and its range is [−π2,π2][-\dfrac\pi2,\dfrac\pi2].
The two endpoints are included, and there are no horizontal asymptotes.
arcsin⁡(13)≈0.33984\arcsin(\dfrac{1}{3})\approx 0.33984 in radians. sin⁡−1(u)\sin^{-1}(u) denotes this inverse here, not 1sin⁡u\dfrac1{\sin u}.
Written-work review:

  • Use the stated definitions and the complete allowed domain.
  • Show all intermediate values and justify signs, restrictions and conclusions.
  • Give every requested result in the stated order and units.

Answer: Domain [−1,1][-1,1]; range [−π2,π2][-\dfrac\pi2,\dfrac\pi2]; value 0.339840.33984.

Inverse identity conjectures and proof

For u=x+34u=x + \dfrac{3}{4}, graph and conjecture formulas for arcsin⁡u+arccos⁡u\arcsin u+\arccos u and arctan⁡u+arctan⁡(1u)\arctan u+\arctan(\frac{1}{u}). Then prove both formulas on their full real domains, handling positive and negative reciprocal inputs separately. Explain the break visible in the second graph.
First inverse sum with closed domain endpointsHorizontal and vertical coordinates use numeric ticks rounded to three decimals. Curves are clipped to the stated window.−3.75−2.55−1.35−0.151.052.25−2−1.2−0.40.41.22LR
Reciprocal inverse sum with open branch endpointsHorizontal and vertical coordinates use numeric ticks rounded to three decimals. Curves are clipped to the stated window.−3.75−2.55−1.35−0.151.052.25−2−1.2−0.40.41.22openopen

First graph: filled endpoints LL and RR are included. Second graph: hollow circles at the two branch ends exclude the reciprocal input zero.

Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

Let u=x+34u=x + \dfrac{3}{4}. The first sum requires −1≤u≤1-1\le u\le1, giving the displayed closed interval. If α=arcsin⁡u\alpha=\arcsin u, then π2−α∈[0,π]\dfrac\pi2-\alpha\in[0,\pi ] and cos⁡(π2−α)=sin⁡α=u\cos(\dfrac\pi2-\alpha)=\sin\alpha=u. Hence arccos⁡u=π2−arcsin⁡u\arccos u=\dfrac\pi2-\arcsin u and the sum is π2\dfrac{\pi }{2}, including endpoints.
For u>0u>0, let β=arctan⁡u∈(0,π2)\beta=\arctan u\in(0,\dfrac\pi2). Its complementary angle has tangent 1u\frac{1}{u} and lies in the arctangent range. Therefore the second sum is π2\dfrac{\pi }{2}.
For u<0u<0, put w=−u>0w=-u>0. Oddness of arctangent gives arctan⁡u+arctan⁡(1u)=−[arctan⁡w+arctan⁡(1w)]=−π2\arctan u+\arctan(\frac{1}{u})=-[\arctan w+\arctan(\frac{1}{w})]=-\dfrac\pi2. At u=0u=0 the reciprocal is undefined, so the two graph pieces are not joined.
Because 11 is positive, u>0u>0 is equivalent to x>−34x>- \dfrac{3}{4}. The numerical graph suggests these identities; the branch argument proves them globally.
Written-work review:

  • Give both complete domains and draw the two disconnected reciprocal-sum pieces.
  • Prove the first identity using complementary principal angles, including endpoints.
  • Prove the positive and negative arctangent cases separately, excluding the zero reciprocal input.

Answer: First sum π2\dfrac{\pi }{2} on −74≤x≤14- \dfrac{7}{4}\le x\le \dfrac{1}{4}; second sum {π2x>−34−π2x<−34\begin{cases}\dfrac{\pi }{2} & x>- \dfrac{3}{4} \\ - \dfrac{\pi }{2} & x<- \dfrac{3}{4}\end{cases}.

Compare compositions globally

Compare both composition orders for the affine input u=x+25u=x + \dfrac{2}{5}: f(x)=sin⁡(arcsin⁡u)f(x)=\sin(\arcsin u) and g(x)=arcsin⁡(sin⁡u)g(x)=\arcsin(\sin u). Draw both actual graphs, state complete domains and ranges, derive a formula valid on every branch of the triangular wave, and explain why blanket cancellation fails.
Restricted line segment and global triangular waveHorizontal and vertical coordinates use numeric ticks rounded to three decimals. Curves are clipped to the stated window.−6.683−4.17−1.6570.8573.375.883−2−1.2−0.40.41.22AB
Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

For f(x)=sin⁡(arcsin⁡(x+25))f(x)=\sin(\arcsin(x + \dfrac{2}{5})), the inner inverse exists only when −1≤x+25≤1-1\le x + \dfrac{2}{5}\le1. Solving gives [−75,35][- \dfrac{7}{5},\dfrac{3}{5}]. There f(x)=x+25f(x)=x + \dfrac{2}{5}, with range [−1,1][-1,1].
For g(x)=arcsin⁡(sin⁡(x+25))g(x)=\arcsin(\sin(x + \dfrac{2}{5})), sine always produces a value in [−1,1][-1,1], so every real input is allowed. The output always belongs to [−π2,π2][-\dfrac\pi2,\dfrac\pi2].
g(x)=(−1)k(u−kπ)g(x)=(-1)^{k}(u-k\pi ) when −π2+kπ≤u≤π2+kπ-\dfrac\pi2+k\pi \le u\le\dfrac\pi2+k\pi , for any integer kk. Adjacent formulas agree at their endpoints, creating the triangular wave.
Increasing x+25x + \dfrac{2}{5} by 2π2\pi repeats the wave; consecutive same-height maxima are exactly 2π2 \pi apart in xx, so this is its least positive period.
The equation g(x)=x+25g(x)=x + \dfrac{2}{5} holds only when −π2≤x+25≤π2-\dfrac\pi2\le x + \dfrac{2}{5}\le\dfrac\pi2. The two compositions are equal on the smaller first domain, but are not the same function globally because their domains differ.
Written-work review:

  • Restrict the first line segment to the inverse domain and keep the second wave defined on all real inputs.
  • Give a branch formula with integer index and correct boundary behavior.
  • Explain principal-range folding and identify the genuine least period.

Answer: First domain [−75,35][- \dfrac{7}{5},\dfrac{3}{5}] and range [−1,1][-1,1]; second domain R\mathbb{R}, range [−π2,π2][-\dfrac\pi2,\dfrac\pi2], period 2π2 \pi .

Inverse reciprocal branches

Use the course branch [0,π2)∪[π,3π2)[0,\dfrac\pi2)\cup[\pi ,\dfrac{3\pi }2) for sec⁡\sec . State the domain and range of its inverse, reflect the restricted original graph to sketch the inverse, identify every horizontal asymptote and endpoint, and evaluate arcsec⁡(2)\operatorname{arcsec}(2), arcsec⁡(−2)\operatorname{arcsec}(-2), arcsec⁡(1)\operatorname{arcsec}(1), arcsec⁡(−1)\operatorname{arcsec}(-1) to five decimals.
Textbook reciprocal-inverse branch conventionHorizontal and vertical coordinates use numeric ticks rounded to three decimals. Curves are clipped to the stated window.−6−3.6−1.21.23.66−0.30.781.862.944.025.1AB
Show the requested reasoning; written work and sketches require teacher review. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

The course convention restricts sec⁡\sec to [0,π2)∪[π,3π2)[0,\dfrac\pi2)\cup[\pi ,\dfrac{3\pi }2). It takes each original function output exactly once. Its inverse therefore has domain (−∞,−1]∪[1,∞)(-\infty,-1]\cup[1,\infty) and range [0,π2)∪[π,3π2)[0,\dfrac\pi2)\cup[\pi ,\dfrac{3\pi }2).
Reflecting the restricted graph across y=xy=x gives the displayed inverse. As input tends to positive infinity, output tends to π2\dfrac{\pi }{2} from below; toward negative infinity it tends to 3π2\dfrac{3 \pi }{2} from below.
arcsec⁡(2)=π3≈1.04720\operatorname{arcsec}(2)=\dfrac{\pi }{3}\approx 1.04720 radians.
arcsec⁡(−2)=4π3≈4.18879\operatorname{arcsec}(-2)=\dfrac{4 \pi }{3}\approx 4.18879 radians.
arcsec⁡(1)=0≈0.00000\operatorname{arcsec}(1)=0\approx 0.00000 radians.
arcsec⁡(−1)=π≈3.14159\operatorname{arcsec}(-1)=\pi \approx 3.14159 radians.
The bracket choices include exactly the finite endpoint values shown. Do not substitute a calculator convention from a different branch without adjustment. The inverse is not the reciprocal of the function.
Written-work review:

  • Use the stated definitions and the complete allowed domain.
  • Show all intermediate values and justify signs, restrictions and conclusions.
  • Give every requested result in the stated order and units.

Answer: Range [0,π2)∪[π,3π2)[0,\dfrac\pi2)\cup[\pi ,\dfrac{3\pi }2); values 1.047201.04720, 4.188794.18879, 0.000000.00000, 3.141593.14159.

15 questions with a worked answer for each.Start practice ↗
Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.

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