Precalculus Flashcards: Constructing Inverse Trigonometric Functions

What this deck covers

This deck focuses on Constructing Inverse Trigonometric Functions, giving you a quick way to review the definitions, rules, and examples that matter most for Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the standard restricted domain for defining $\arccos(x)$ as an inverse of $\cos(x)$?

Answer: $\left[0,\pi\right]$. On this interval, cosine is strictly decreasing from 1 to -1.

Flashcard 2: Identify whether $\tan(x)$ is increasing or decreasing on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Answer: Increasing on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Derivative $\sec^2(x)>0$ on this interval confirms monotonic increase.

Flashcard 3: What is the relationship between the graphs of $y=f(x)$ and $y=f^{-1}(x)$?

Answer: They are reflections across the line $y=x$. Inverse functions are symmetric about the line $y=x$.

Flashcard 4: Identify the principal-value identity for $\sin(\arcsin(x))$.

Answer: $\sin(\arcsin(x))=x$ for $x\in\left[-1,1\right]$. Applying sine to its inverse returns the original value.

Flashcard 5: What identity expresses the inverse relationship on the domain of $\arccos$?

Answer: $\cos(\arccos(x))=x$ for $x\in\left[-1,1\right]$. Composing a function with its inverse yields the identity function.

Flashcard 6: What is the simplified value of $\arccos\left(\cos\left(\frac{5\pi}{3}\right)\right)$?

Answer: $\frac{\pi}{3}$. $\cos(\frac{5\pi}{3})=\frac{1}{2}$, and $\arccos(\frac{1}{2})=\frac{\pi}{3}$ in range.

Flashcard 7: Identify the restricted interval where $\tan(x)$ is strictly increasing and invertible.

Answer: $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Tangent increases continuously without vertical asymptotes here.

Flashcard 8: What does it mean for a function to be one-to-one in terms of the horizontal line test?

Answer: Every horizontal line intersects the graph at most once. This test verifies that each y-value corresponds to at most one x-value.

Flashcard 9: What restricted domain is used to define $\arccos(x)$ as the inverse of $\cos(x)$?

Answer: $\left[0,\pi\right]$. On this interval, cosine decreases from 1 to -1, making it one-to-one.

Flashcard 10: Identify whether $\sin(x)$ is increasing or decreasing on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Answer: Increasing on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Derivative $\cos(x)>0$ on this interval confirms monotonic increase.

Flashcard 11: Identify the restricted interval where $\cos(x)$ is strictly decreasing and invertible.

Answer: $\left[0,\pi\right]$. Cosine decreases monotonically from 1 to -1 on this interval.

Flashcard 12: What is the simplified value of $\arctan\left(\tan\left(\frac{3\pi}{4}\right)\right)$?

Answer: $-\frac{\pi}{4}$. $\tan(\frac{3\pi}{4})=-1$, and $\arctan(-1)=-\frac{\pi}{4}$ in range.

Flashcard 13: Identify whether $\cos(x)$ is increasing or decreasing on $\left[0,\pi\right]$.

Answer: Decreasing on $\left[0,\pi\right]$. Derivative $-\sin(x)<0$ on this interval confirms monotonic decrease.

Flashcard 14: Identify the principal-value identity for $\cos(\arccos(x))$.

Answer: $\cos(\arccos(x))=x$ for $x\in\left[-1,1\right]$. Applying cosine to its inverse returns the original value.

Flashcard 15: What is the range of $\arctan(x)$?

Answer: $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. This matches the restricted domain of tangent used to define its inverse.

Flashcard 16: What is the range of $\arccos(x)$?

Answer: $\left[0,\pi\right]$. This matches the restricted domain of cosine used to define its inverse.

Flashcard 17: What identity expresses the inverse relationship on the domain of $\arctan$?

Answer: $\tan(\arctan(x))=x$ for $x\in\left(-\infty,\infty\right)$. Composing a function with its inverse yields the identity function.

Flashcard 18: What is the domain of $\arcsin(x)$?

Answer: $\left[-1,1\right]$. Since sine's range is $[-1,1]$, this becomes arcsine's domain.

Flashcard 19: What is the standard restricted domain for defining $\arcsin(x)$ as an inverse of $\sin(x)$?

Answer: $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. On this interval, sine is strictly increasing from -1 to 1.

Flashcard 20: What restricted domain is used to define $\arctan(x)$ as the inverse of $\tan(x)$?

Answer: $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Tangent is strictly increasing on this interval without vertical asymptotes.

Flashcard 21: Identify the restricted interval where $\sin(x)$ is strictly increasing and invertible.

Answer: $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Sine increases monotonically from -1 to 1 on this interval.

Flashcard 22: What is the standard restricted domain for defining $\arctan(x)$ as an inverse of $\tan(x)$?

Answer: $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. On this interval, tangent is strictly increasing through all real values.

Flashcard 23: What property must a trigonometric function have on a restricted domain to have an inverse?

Answer: It must be one-to-one (strictly increasing or strictly decreasing). This ensures each output maps to exactly one input, making the function invertible.

Flashcard 24: What is the domain of $\arccos(x)$?

Answer: $\left[-1,1\right]$. Since cosine's range is $[-1,1]$, this becomes arccosine's domain.

Flashcard 25: Evaluate $\arccos\left(-\frac{1}{2}\right)$ using the principal range of $\arccos$.

Answer: $\frac{2\pi}{3}$. Since $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$ and $\frac{2\pi}{3} \in [0, \pi]$.

Flashcard 26: Which restricted interval makes $\sin(x)$ one-to-one: $\left[0,2\pi\right]$ or $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$?

Answer: $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Only the second interval has sine strictly monotonic throughout.

Flashcard 27: What restricted domain is used to define $\arcsin(x)$ as the inverse of $\sin(x)$?

Answer: $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. On this interval, sine increases from -1 to 1, passing the horizontal line test.

Flashcard 28: Evaluate $\arcsin\left(\frac{1}{2}\right)$ using the principal range of $\arcsin$.

Answer: $\frac{\pi}{6}$. Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Flashcard 29: What is the range of $\arcsin(x)$?

Answer: $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. This matches the restricted domain of sine used to define its inverse.

Flashcard 30: What is the simplified value of $\arcsin\left(\sin\left(\frac{3\pi}{4}\right)\right)$?

Answer: $\frac{\pi}{4}$. $\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}$, and $\arcsin(\frac{\sqrt{2}}{2})=\frac{\pi}{4}$ in range.

Flashcard 31: Identify the principal-value identity for $\tan(\arctan(x))$.

Answer: $\tan(\arctan(x))=x$ for $x\in\left(-\infty,\infty\right)$. Applying tangent to its inverse returns the original value.

Flashcard 32: What identity expresses the inverse relationship on the domain of $\arcsin$?

Answer: $\sin(\arcsin(x))=x$ for $x\in\left[-1,1\right]$. Composing a function with its inverse yields the identity function.

Flashcard 33: What property must a trigonometric function have on a restricted domain to have an inverse there?

Answer: It must be one-to-one (strictly increasing or strictly decreasing). This ensures each input maps to exactly one output, allowing inverse construction.