Inverse Trigonometric Functions and Their Applications

Inverse Trigonometric Functions

Introduction to Inverse Functions

Inverse functions allow us to 'undo' the action of a function. In trigonometry, inverse trigonometric functions are used to find angles when given trigonometric values.

• Horizontal Line Test: A function has an inverse if and only if every horizontal line intersects its graph at most once (i.e., the function is one-to-one).

• Graphical Relationship: If the point (a, b) is on the graph of a function f, then (b, a) is on the graph of its inverse, denoted as $f^{-1}$.

• Reflection: The graph of $f^{-1}$ is a reflection of the graph of f about the line $y = x$.

Inverse Sine Function

Definition and Properties

The inverse sine function, denoted as $\sin^{-1} x$ or $\arcsin x$, is the inverse of the sine function restricted to the domain $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$.

• Domain: $-1 \leq x \leq 1$

• Range: $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

• Interpretation: $y = \sin^{-1} x$ means $\sin y = x$ for $y$ in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Graphing the Inverse Sine Function

• To graph $y = \sin^{-1} x$, take points from the graph of $y = \sin x$ (restricted to $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$) and reverse the order of the coordinates.

• Alternatively, reflect the graph of the restricted sine function about the line $y = x$.

Table: Common Values of $\sin^{-1} x$

$x$	$-1$	$-\frac{\sqrt{3}}{2}$	$-\frac{\sqrt{2}}{2}$	$-\frac{1}{2}$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
$\sin^{-1} x$	$-\frac{\pi}{2}$	$-\frac{\pi}{3}$	$-\frac{\pi}{4}$	$-\frac{\pi}{6}$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$

Example: Finding the Exact Value of an Inverse Sine

• Problem: Find the exact value of $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.

• Step 1: Let $\theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.

• Step 2: Rewrite as $\sin \theta = \frac{\sqrt{3}}{2}$, where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

• Step 3: From the table, the angle in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ whose sine is $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$.

• Answer: $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

Applications

• Inverse trigonometric functions are used to solve for angles in right triangles when side lengths are known.

• They are also essential in solving equations involving trigonometric expressions and in modeling periodic phenomena.

Additional info: The notes continue with similar treatments for the inverse cosine and inverse tangent functions, including their definitions, domains, ranges, graphs, and example problems. These are standard topics in a Precalculus course.