Comprehensive Study Notes on Inverse Trigonometric Functions

Fundamentals of Inverse Trigonometric Functions

• Terminology and Notation Interchanging:

  • The inverse sine function can be written as $\sin^{-1}(x)$ or $\arcsin(x)$.
  • The prefix "arc" refers to the arc length of a circle.
  • On a unit circle, the angle $\theta$ in radians is defined by the arc length $s$ divided by the radius $r$:     $\theta = \frac{s}{r}$
  • In a standard trigonometric function, the input is an angle $\theta$ (in degrees or radians) or arc length $s$, and the output is a trigonometric ratio.
  • In an inverse trigonometric function, the inputs and outputs are inverted:
  • Standard Trigonometric Function: Input = Angle / Arc Length $\rightarrow$ Output = Ratio
  • Inverse Trigonometric Function: Input = Ratio $\rightarrow$ Output = Angle / Arc Length
  • The notation $\arcsin(x)$, $\arccos(x)$, and $\arctan(x)$ is used interchangeably with $\sin^{-1}(x)$, $\cos^{-1}(x)$, and $\tan^{-1}(x)$.

• The Necessity of Restricted Domains:

  • A relation must assign exactly one output for every input to qualify as a function.
  • For the standard sine function, multiple angles yield the exact same output ratio. For example:     $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$$\sin\left(\frac{9\pi}{4}\right) = \frac{\sqrt{2}}{2}$$\sin\left(\frac{11\pi}{4}\right) = \frac{\sqrt{2}}{2}$
  • If the inverse mapping were evaluated without domain restrictions, an input ratio of $\frac{\sqrt{2}}{2}$ would yield infinitely many outputs ($\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{9\pi}{4}$, $\frac{11\pi}{4}$, etc.), which violates the definition of a function.
  • To make inverse trigonometric relations true functions, restrictions must be placed on the domains of the original trigonometric functions, limiting their outputs (the range of the inverse) to a single contiguous interval containing a unique angle for every valid ratio.
  • By mathematical convention, domain restrictions are centered as close to $0$ as possible, favoring positive angles while maintaining graph connectivity.

Inverse Sine Function ($\arcsin(x)$ or $\sin^{-1}(x)$)

• Domain and Range Specifications:

  • For the restricted standard sine function $y = \sin(x)$, the restricted domain is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and the range is $[-1, 1]$.
  • Swapping inputs and outputs for $y = \arcsin(x)$ yields:
  • Domain: $[-1, 1]$
  • Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$

• Quadrant Mapping for Inverse Sine:

  • Positive ratios ($x \in [0, 1]$) map to Quadrant I angles ($y \in [0, \frac{\pi}{2}]$).
  • Negative ratios ($x \in [-1, 0)$) map to Quadrant IV angles ($y \in [-\frac{\pi}{2}, 0)$).

• Algebraic Property (Odd Function):

  • The inverse sine function is an odd function, meaning:     $\arcsin(-x) = -\arcsin(x)$
  • Algebraic Verification:
  • Let $y = \sin^{-1}(x)$, which implies $\sin(y) = x$.
  • Let $\theta = \sin^{-1}(-x)$, which implies $\sin(\theta) = -x$.
  • Substituting $x = \sin(y)$ into the equation gives $\sin(\theta) = -\sin(y)$.
  • Since standard sine is an odd function, $-\sin(y) = \sin(-y)$, giving $\sin(\theta) = \sin(-y)$.
  • Taking the inverse sine of both sides yields $\theta = -y$.
  • Replacing $\theta$ and $y$ with their definitions yields $\sin^{-1}(-x) = -\sin^{-1}(x)$.

• Step-by-Step Evaluated Examples:

  • Example 1: Find the exact value of $y = \arcsin\left(\frac{\sqrt{3}}{2}\right)$.
  • Rewrite as a standard sine equation: $\sin(y) = \frac{\sqrt{3}}{2}$, where $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
  • Since the ratio $\frac{\sqrt{3}}{2}$ is positive, $y$ must lie in Quadrant I ($[0, \frac{\pi}{2}]$).
  • On the unit circle, the point with a $y$-coordinate of $\frac{\sqrt{3}}{2}$ in Quadrant I is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$, corresponding to an angle of $\frac{\pi}{3}$.
  • Alternatively, using a right triangle with an opposite side of $\sqrt{3}$ and a hypotenuse of $2$, the angle opposite to $\sqrt{3}$ is $60^\circ$, which equals $\frac{\pi}{3}$ radians.
  • Output: $y = \frac{\pi}{3}$.
  • Example 2: Find the exact value of $y = \sin^{-1}\left(-\frac{1}{2}\right)$.
  • Rewrite as a standard sine equation: $\sin(y) = -\frac{1}{2}$, where $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
  • Since the ratio $-\frac{1}{2}$ is negative, $y$ must lie in Quadrant IV ($[-\frac{\pi}{2}, 0]$).
  • Using the reference angle method: $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Applying the odd function property gives:       $\sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}$
  • Unit circle point in Quadrant IV: $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ at angle $-\frac{\pi}{6}$.
  • Output: $y = -\frac{\pi}{6}$.
  • Example 3: Find the exact value of $y = \sin^{-1}(-3)$.
  • Rewrite as a standard sine equation: $\sin(y) = -3$.
  • Evaluate the input against the domain restriction of inverse sine ($[-1, 1]$).
  • Because $-3 < -1$, the ratio $-3$ falls outside the domain of $\sin^{-1}(x)$.
  • Output: Does Not Exist (DNE).

Inverse Cosine Function ($\arccos(x)$ or $\cos^{-1}(x)$)

• Domain and Range Specifications:

  • Key standard cosine values: $\cos(0) = 1$, $\cos\left(\frac{\pi}{2}\right) = 0$, and $\cos(\pi) = -1$.
  • To create a one-to-one function that captures all standard output ratios from $1$ to $-1$ without repeating values, the cosine function is restricted to the interval $[0, \pi]$.
  • Inverting inputs and outputs for $y = \arccos(x)$ yields:
  • Domain: $[-1, 1]$
  • Range: $[0, \pi]$

• Quadrant Mapping for Inverse Cosine:

  • Positive ratios ($x \in [0, 1]$) map to Quadrant I angles ($y \in [0, \frac{\pi}{2}]$).
  • Negative ratios ($x \in [-1, 0)$) map to Quadrant II angles ($y \in (\frac{\pi}{2}, \pi]$).
  • Critical Difference: The range of $\arccos(x)$ contains no negative angles. Negative ratios output angles in Quadrant II, unlike $\arcsin(x)$ which uses Quadrant IV negative angles.

• Step-by-Step Evaluated Examples:

  • Example 1: Find the exact value of $y = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$.
  • Rewrite as a standard cosine equation: $\cos(y) = \frac{\sqrt{2}}{2}$, where $y \in [0, \pi]$.
  • Since the ratio is positive, $y$ lies in Quadrant I.
  • On the unit circle, the point $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ occurs at an angle of $\frac{\pi}{4}$ (or $45^\circ$).
  • Output: $y = \frac{\pi}{4}$.
  • Example 2: Find the exact value of $y = \arccos\left(-\frac{1}{2}\right)$.
  • Rewrite as a standard cosine equation: $\cos(y) = -\frac{1}{2}$, where $y \in [0, \pi]$.
  • Since the ratio is negative, $y$ must lie in Quadrant II ($[\frac{\pi}{2}, \pi]$).
  • Determine the reference angle $\theta'$ in Quadrant I where $\cos(\theta') = \frac{1}{2}$.       $\theta' = \frac{\pi}{3}$
  • To find the Quadrant II angle $y$, subtract the reference angle from $\pi$:       $y = \pi - \theta' = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
  • Unit circle check: $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ at angle $\frac{2\pi}{3}$.
  • Output: $y = \frac{2\pi}{3}$.

Inverse Tangent Function ($\arctan(x)$ or $\tan^{-1}(x)$)

• Domain and Range Specifications:

  • The standard tangent function $y = \tan(x)$ has vertical asymptotes at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.
  • As $x \rightarrow \left(\frac{\pi}{2}\right)^{-}$, $\tan(x) \rightarrow \infty$. As $x \rightarrow \left(-\frac{\pi}{2}\right)^{+}$, $\tan(x) \rightarrow -\infty$.
  • Restricting the tangent function to its central branch between its asymptotes, $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, covers all real numbers smoothly.
  • Inverting inputs and outputs for $y = \arctan(x)$ yields:
  • Domain: $(-\infty, \infty)$ (All real numbers)
  • Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
  • Note that open intervals (parentheses) are used because $\tan\left(\pm \frac{\pi}{2}\right)$ is undefined.

• Graphical Characteristics of Inverse Tangent:

  • The vertical asymptotes of the tangent function transform into horizontal asymptotes for the inverse tangent graph:     $y = \frac{\pi}{2}$$y = -\frac{\pi}{2}$
  • Inverse tangent is an odd function:     $\arctan(-x) = -\arctan(x)$

• Step-by-Step Evaluated Examples:

  • Example 1: Find the exact value of $y = \tan^{-1}(0)$.
  • Rewrite as a standard tangent equation: $\tan(y) = 0$, where $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
  • Using quotient identities: $\tan(y) = \frac{\sin(y)}{\cos(y)} = 0 \implies \sin(y) = 0$
  • Within $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\sin(y) = 0$ occurs at $y = 0$.
  • Output: $y = 0$
  • Example 2: Find the exact value of $y = \arctan(-\sqrt{3})$.
  • Rewrite as a standard tangent equation: $\tan(y) = -\sqrt{3}$, where $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
  • Using odd function properties: $\arctan(-\sqrt{3}) = -\arctan(\sqrt{3})$.
  • Draw a reference triangle in Quadrant I where $\tan(\theta') = \sqrt{3} = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{3}}{1}$.
  • Hypotenuse = $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$.
  • The angle opposite to $\sqrt{3}$ is $60^\circ$ or $\frac{\pi}{3}$.
  • Applying the negative sign for Quadrant IV gives $y = -\frac{\pi}{3}$.
  • Unit circle check: $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \implies \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3}$.
  • Output: $y = -\frac{\pi}{3}$.
  • Example 3: Find the exact value of $y = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$.
  • Rewrite as a standard tangent equation: $\tan(y) = \frac{\sqrt{3}}{3}$, where $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
  • Simplify the ratio by rationalizing in reverse:       $\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{1}{\sqrt{3}}$
  • Draw a reference triangle with opposite side $1$ and adjacent side $\sqrt{3}$.
  • The angle opposite to side $1$ in a $30^\circ-60^\circ-90^\circ$ triangle is $30^\circ$ or $\frac{\pi}{6}$.
  • Unit circle check: $\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.
  • Output: $y = \frac{\pi}{6}$.

Inverse Reciprocal Trigonometric Functions

• Reciprocal Identity Context:

  • $\csc(\theta) = \frac{1}{\sin(\theta)}$
  • $\sec(\theta) = \frac{1}{\cos(\theta)}$
  • $\cot(\theta) = \frac{1}{\tan(\theta)}$

• Inverse Cosecant Function ($\csc^{-1}(x)$ or $\text{arccsc}(x)$):

  • Domain: $(-\infty, -1] \cup [1, \infty)$ (or $|x| \ge 1$). Derived directly from the range of $\csc(\theta)$.
  • Range: $\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$. Excludes $y = 0$ because $\sin(0) = 0$, which makes $\csc(0) = \frac{1}{0}$ undefined.
  • Example A: Find the exact value of $y = \csc^{-1}\left(\frac{1}{2}\right)$.
  • Rewrite equation: $\csc(y) = \frac{1}{2} \implies \sin(y) = 2$
  • Since $\sin(y)$ cannot exceed $1$, and $\frac{1}{2}$ is outside the domain of $|x| \ge 1$, this value cannot be calculated.
  • Output: Does Not Exist (DNE).
  • Example B: Find the exact value of $y = \csc^{-1}(2)$.
  • Rewrite equation: $\csc(y) = 2 \implies \sin(y) = \frac{1}{2}$
  • For positive ratios, $y$ lies in Quadrant I ($(0, \frac{\pi}{2}]$).
  • $\sin(y) = \frac{1}{2} \implies y = \frac{\pi}{6}$.
  • Output: $y = \frac{\pi}{6}$.

• Inverse Secant Function ($\sec^{-1}(x)$ or $\text{arcsec}(x)$):

  • Domain: $(-\infty, -1] \cup [1, \infty)$ (or $|x| \ge 1$).
  • Range: $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$. Excludes $y = \frac{\pi}{2}$ because $\cos\left(\frac{\pi}{2}\right) = 0$, making $\sec\left(\frac{\pi}{2}\right)$ undefined.
  • Quadrant Mapping: Positive ratios map to Quadrant I ($[0, \frac{\pi}{2})$); negative ratios map to Quadrant II ($(\frac{\pi}{2}, \pi]$).

• Inverse Cotangent Function ($\cot^{-1}(x)$ or $\text{arccot}(x)$):

  • Domain: $(-\infty, \infty)$ (All real numbers).
  • Range: $(0, \pi)$. Excludes $0$ and $\pi$ because $\tan(0) = 0$ and $\tan(\pi) = 0$, making $\cot(0)$ and $\cot(\pi)$ undefined.
  • Quadrant Mapping: Positive ratios map to Quadrant I ($(0, \frac{\pi}{2})$); negative ratios map to Quadrant II ($(\frac{\pi}{2}, \pi)$).
  • Example: Find the exact value of $y = \cot^{-1}(-\sqrt{3})$.
  • Rewrite equation: $\cot(y) = -\sqrt{3}$, where $y \in (0, \pi)$.
  • Since the ratio is negative, $y$ must lie in Quadrant II ($(\frac{\pi}{2}, \pi)$).
  • Set up quotient definition: $\cot(y) = \frac{\cos(y)}{\sin(y)} = \frac{\text{adjacent}}{\text{opposite}} = -\sqrt{3}$.
  • Find the reference angle $\theta'$ in Quadrant I where $\cot(\theta') = \sqrt{3} \implies \tan(\theta') = \frac{1}{\sqrt{3}} \implies \theta' = \frac{\pi}{6}$.
  • Calculate Quadrant II angle $y$ using reference angle $\theta'$:       $y = \pi - \theta' = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
  • Unit circle check: point $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \implies \cot\left(\frac{5\pi}{6}\right) = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3}$.
  • Output: $y = \frac{5\pi}{6}$.

Summary of Domains and Ranges for Inverse Trigonometric Functions

• Comprehensive Reference Table:
  • $\arcsin(x)$, Domain: $[-1, 1]$, Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$
  • $\arccos(x)$, Domain: $[-1, 1]$, Range: $[0, \pi]$
  • $\arctan(x)$, Domain: $(-\infty, \infty)$, Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
  • $\csc^{-1}(x)$, Domain: $(-\infty, -1] \cup [1, \infty)$, Range: $\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$
  • $\sec^{-1}(x)$, Domain: $(-\infty, -1] \cup [1, \infty)$, Range: $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$
  • $\cot^{-1}(x)$, Domain: $(-\infty, \infty)$, Range: $(0, \pi)$

Classroom Dialogue and Student Interactions

• Inquiry on Terminology:

  • Student Question: Why is the term "arc sine" used instead of inverse sine?
  • Explanation Provided: On the unit circle, $\theta = \frac{s}{r}$, where $s$ represents arc length. Standard sine inputs an angle or arc length to output a ratio. Inverse sine inputs a ratio to output the corresponding arc length or angle, hence the term "arc sine".

• Inquiry on Mapping and Functions:

  • Student Interaction (Josh): Shook head when asked if mapping $\frac{\sqrt{2}}{2}$ back to $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ forms a function.
  • Josh's Reasoning: Pointed out that angles were being used inappropriately or mapped to multiple places.
  • Clarification: A single input mapping to multiple outputs (e.g., $\frac{\sqrt{2}}{2} \rightarrow \frac{\pi}{4}$ and $\frac{\sqrt{2}}{2} \rightarrow \frac{3\pi}{4}$) violates the definition of a function. The domain must be restricted so each input yields exactly one unique output.

• Inquiry on Negative Quadrant Mapping:

  • Student Interaction (Amber): Expressed confusion regarding why output angles are negative for some inverse functions and positive for others.
  • Clarification: For $\arcsin(x)$ and $\arctan(x)$, negative ratio inputs return negative angles in Quadrant IV ($[-\frac{\pi}{2}, 0)$ and $\left(-\frac{\pi}{2}, 0\right)$). For $\arccos(x)$ and $\cot^{-1}(x)$, negative ratio inputs return positive angles in Quadrant II ($(\frac{\pi}{2}, \pi]$ and $\left(\frac{\pi}{2}, \pi\right)$) because their ranges are restricted to $[0, \pi]$ and (0, \n\pi).

• Course Administration Note:

  • Written Assignment Deadline: The written assignment deadline originally scheduled for Thursday was extended to Friday.