Inverse Trigonometric and Hyperbolic Functions Study of Hyperbolic Functions Study Notes

Evaluating Inverse Trigonometric Functions

• Overview of Evaluation: Inverse trigonometric functions, such as arcsine ($\sin^{-1}$), are evaluated using methods accepted in textbooks and by standard calculators. While some exceptions exist, the primary method involves working backward from known exact values.
• Cancellation Formulas: These formulas define the Relationship between a function $f$ and its inverse $f^{-1}$.
  • $f(f^{-1}(x)) = x$
  • $f^{-1}(f(x)) = x$
  • Essentially, the function and its inverse "cancel out."
• Working Backwards from Exact Values: If the exact value of a trigonometric function is known (e.g., from Workshop 1 formula sheets), the cancellation formula can be applied to both sides to find the inverse.
  • Example 1: Since $\sin(0) = 0$, applying $\sin^{-1}$ to both sides yields $\sin^{-1}(\sin(0)) = \sin^{-1}(0)$, which simplifies to $0 = \sin^{-1}(0)$.
  • Example 2: Since $\sin(\frac{\pi}{2}) = 1$, applying the same logic results in $\frac{\pi}{2} = \sin^{-1}(1)$.
• Preference for Exact Answers: Students are encouraged to provide exact answers whenever possible, especially when working with known exact values, rather than relying solely on decimal outputs from a calculator.

The Sine Inverse Function ($\sin^{-1}$)

• Restricted Domain and Range: To define an inverse for the periodic sine function, its domain must be restricted so the function becomes one-to-one.
  • The standard restriction for sine is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
  • Consequently, the range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, which corresponds to the first and fourth quadrants on the unit circle.
• Unit Circle Interpretation:
  • $-\frac{\pi}{2}$ represents the angle moving clockwise to the bottom of the circle.
  • $\frac{\pi}{2}$ represents the top of the circle.
  • Calculations typically fall within these two quadrants. If a problem requires an answer in quadrants two or three, an adjustment based on the sign of the function is necessary (a process used frequently in complex numbers).
• Formal Cancellation Definitions for Sine:
  • $\sin^{-1}(\sin(x)) = x$ where $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
  • $\sin(\sin^{-1}(x)) = x$ where $x \in [-1, 1]$.
• Evaluation Examples and Validity Checks:
  • Example A: $\sin^{-1}(\sin(\frac{\pi}{6})) = \frac{\pi}{6}$ because $\frac{\pi}{6}$ is within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
  • Example B: $\sin^{-1}(\sin(-\frac{\pi}{3})) = -\frac{\pi}{3}$ because $-\frac{\pi}{3}$ is within the valid interval.
  • Example C: $\sin(\sin^{-1}(0.3)) = 0.3$ because $0.3$ is within the interval $[-1, 1]$.
  • Example D (Nonexistent case): $\sin(\sin^{-1}(1.5))$ does not exist. Although a literal application of cancellation suggests $1.5$, the value $1.5$ is outside the domain of sine ($[-1, 1]$). Sine cannot output a value greater than $1$.
  • Example E (Periodic adjustment): $\sin^{-1}(\sin(\frac{5\pi}{6}))$. Since $\frac{5\pi}{6}$ is outside the restricted range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, one must find the reference angle. The reference angle for $\frac{5\pi}{6}$ is $\frac{\pi}{6}$. Because sine is positive in both the first and second quadrants, $\sin(\frac{5\pi}{6}) = \sin(\frac{\pi}{6})$. Thus, $\sin^{-1}(\sin(\frac{\pi}{6})) = \frac{\pi}{6}$.
• Determining the Domain of an Inverse Function:
  • To find the domain of $f(x) = \sin^{-1}(x - 3)$, set the argument within the restricted range: $-1 \le x - 3 \le 1$.
  • Adding $3$ to all sides: $2 \le x \le 4$.
  • Domain Notation: The domain can be written as the set of all $x$ such that $2 \le x \le 4$. In interval notation, this is $[2, 4]$. Symbols used include the colon ($:$) or vertical bar ($|$) to mean "such that."

The Cosine Inverse Function ($\cos^{-1}$, arccos)

• Definition and Restriction: Like sine, the cosine function is periodic and needs a restricted domain to have an inverse. The accepted restriction is $[0, \pi]$.
• Domain and Range:
  • Domain: $[-1, 1]$.
  • Range: $[0, \pi]$ (Quadrants one and two).
• Meaning of the Function: The notation $y = \cos^{-1}(x)$ implies that $y$ is the angle whose cosine is $x$ (i.e., $\cos(y) = x$).
• Notation Warning: The $-1$ in $\cos^{-1}(x)$ is a superscript denoting an inverse function, not an exponent.
  • $\cos^{-1}(x) \neq (\cos(x))^{-1}$.
  • $(\cos(x))^{-1}$ or $\frac{1}{\cos(x)}$ refers to the reciprocal (secant).
• Graphical Representation: The graph of $\cos^{-1}(x)$ is obtained by sketching the restricted cosine function and reflecting it across the line $y = x$.
• Exact Value Example: Since $\cos(0) = 1$, it follows that $\cos^{-1}(1) = 0$.

The Tangent Inverse Function ($\tan^{-1}$, arctan)

• Restriction and Intervals: The tangent function is restricted to the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. Notice the use of round brackets, meaning the endpoints are excluded ($-\frac{\pi}{2} < x < \frac{\pi}{2}$).
• Domain and Range:
  • Domain: $(-\infty, \infty)$ (all real numbers).
  • Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$.
• Horizontal Asymptotes: Due to the vertical asymptotes of the tangent function, the inverse tangent function possesses horizontal asymptotes:
  • $\lim_{x \rightarrow \infty} \tan^{-1}(x) = \frac{\pi}{2}$
  • $\lim_{x \rightarrow -\infty} \tan^{-1}(x) = -\frac{\pi}{2}$
• Graph: The curve approaches these asymptotes but never crosses them. It is generated by reflecting the restricted tangent function about the line $y = x$.

Cosecant, Secant, and Cotangent Inverse Functions

• Lack of Universal Agreement: There is no universal agreement on the specific domain restrictions for $\csc^{-1}$, $\sec^{-1}$, and $\cot^{-1}$.
• Rationale for Choice: Restrictions are often chosen to make the derivation of the derivative easier. Different sources, such as the course eBook, MATLAB, or specific authors, may use different definitions.
• Adopted Definitions for this Course:
  • Inverse Cosecant ($\csc^{-1}$): Defined on the union of intervals $(0, \frac{\pi}{2}] \cup (\pi, \frac{3\pi}{2}]$. This targets the first and third quadrants.
    • Domain: $|x| \ge 1$ (meaning $x \ge 1$ or $x \le -1$).
  • Inverse Secant ($\sec^{-1}$): Defined on $[0, \frac{\pi}{2}) \cup [\pi, \frac{3\pi}{2})$ (Quadrants one and three).
  • Note the use of square brackets for inclusive values and round brackets for exclusive values (e.g., $\sec^{-1}$ excludes $\frac{\pi}{2}$ because secant is undefined there).
• Software Variations: Computational tools like MATLAB utilize different internal definitions for these functions. This is explored further in Lab 3 or Lab 4 and Workshop 2.

Composite Functions and Reference Triangles

• Composition Definition: A composite function involves a function within a function, denoted as $f(g(x))$ or $(f \circ g)(x)$. This is performed by substituting the function $g(x)$ everywhere there is an $x$ in function $f$.
• Technique: Reference Triangles: To evaluate compositions involving a trigonometric function and an inverse trigonometric function (e.g., $\cos(\sin^{-1}(\frac{\sqrt{3}}{2}))$), treat the inverse function as an angle.
• Example 1: Evaluating $\cos(\sin^{-1}(\frac{\sqrt{3}}{2}))$:
  1. Let $\alpha = \sin^{-1}(\frac{\sqrt{3}}{2})$.
  2. This implies $\sin(\alpha) = \frac{\sqrt{3}}{2}$.
  3. Using a right-angled triangle where $\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}}$, label the opposite side $\sqrt{3}$ and the hypotenuse $2$.
  4. Use the Pythagorean theorem to find the adjacent side: $\sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4 - 3} = 1$.
  5. The original problem now becomes finding $\cos(\alpha)$.
  6. $\cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2}$.
• Example 2: Evaluating $\sec(\tan^{-1}(\frac{x}{3}))$:
  1. Let $\beta = \tan^{-1}(\frac{x}{3})$, implying $\tan(\beta) = \frac{x}{3}$.
  2. Set up a reference triangle where the opposite side is $x$ and the adjacent side is $3$.
  3. The hypotenuse is $\sqrt{x^2 + 3^2} = \sqrt{x^2 + 9}$.
  4. The problem is to find $\sec(\beta)$, which is $\frac{1}{\cos(\beta)}$.
  5. Since $\cos(\beta) = \frac{3}{\sqrt{x^2 + 9}}$, it follows that $\sec(\beta) = \frac{\sqrt{x^2 + 9}}{3}$.

Introduction to Hyperbolic Functions

• Definition and Utility: Hyperbolic functions and their inverses are crucial in calculus, particularly in engineering.
• Applications:
  • Integration: They appear frequently in integral formulas and the evaluation of specific integrals.
  • Modeling: They model physical phenomena such as a hanging wire, known as a catenary, which follows a hyperbolic shape.

Questions & Discussion

• Audience Exchange: During the demonstration of reference triangles, the speaker clarified the illegibility of their handwriting and confirmed the step-by-step logic of the Pythagorean theorem application.
• Student Question: "Are we going to type off?"
• Instructor Response: The instructor continued the lecture, transitioning into hyperbolic functions and noting that most of the detailed definitions for inverse secant and cosecant are provided on the formula sheets for the unit, so memorization is not required as long as pupils know how to apply them accurately.