Inverse Trig Foundations and Domain Restrictions

Foundational

Master why trigonometric functions require domain restrictions to form invertible functions.

Inverse Trig Foundations and Domain Restrictions

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What test must a function pass on a given interval to guarantee the existence of an inverse function?

The horizontal line test. It ensures that every horizontal line intersects the graph at most once, meaning the function is one-to-one (injective).

What are the domain and range of the inverse sine function $y = \arcsin(x)$?

Graph of y equals arcsin of x on the Cartesian plane, illustrating the restricted domain from negative 1 to 1 and range from negative pi over 2 to pi over 2, with endpoints at (-1, -pi/2) and (1, pi/2) and an inflection point at the origin.

Domain: $[-1, 1]$ Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

What standard restricted domain is chosen for $y = \cos(x)$ so that it possesses an inverse?

Graph of the cosine function highlighting the restricted domain from 0 to pi in bold, showing a strictly decreasing curve from 1 to -1 that satisfies the horizontal line test to define the inverse cosine.

$$ [0, \pi] $$ On this interval, $\cos(x)$ is strictly decreasing and covers all output values from $1$ to $-1$ exactly once.

Why is the restricted domain for $y = \cos(x)$ chosen as $[0, \pi]$ rather than $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$?

On $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, cosine is symmetric about the $y$-axis (even), meaning $\cos(-x) = \cos(x)$. It fails the horizontal line test and only covers $[0, 1]$.

What are the domain and range of the inverse cosine function $y = \arccos(x)$?

Graph of y equals arccos(x) showing a decreasing curve on the domain from negative 1 to 1 and range from 0 to pi, highlighting key points at (-1, pi), the y-intercept at (0, pi/2), and (1, 0).

Domain: $[-1, 1]$ Range: $[0, \pi]$

What restricted domain is chosen for $y = \tan(x)$ to define $y = \arctan(x)$?

$$ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $$ The endpoints are excluded because vertical asymptotes occur at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.

What are the domain and range of the inverse tangent function $y = \arctan(x)$?

Graph of the inverse tangent function y = arctan(x), passing through the origin with horizontal asymptotes at y = pi/2 and y = -pi/2, illustrating a domain of all real numbers and range from -pi/2 to pi/2.

Domain: $(-\infty, \infty)$ Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

In which quadrants on the unit circle do the principal output angles of $y = \arccos(x)$ lie?

Unit circle highlighting the top half across Quadrants I and II from 0 to pi, representing the restricted range of the inverse cosine function.

Quadrant I for positive inputs $(0 \le x \le 1)$ and Quadrant II for negative inputs $(-1 \le x < 0)$.

In which quadrants on the unit circle do the principal output angles of $y = \arctan(x)$ lie?

Quadrant I for positive inputs $(x > 0)$ and Quadrant IV as negative angles $\left(-\frac{\pi}{2}, 0\right)$ for negative inputs $(x < 0)$.

What is the crucial mathematical difference between $\sin^{-1}(x)$ and $(\sin(x))^{-1}$?

$\sin^{-1}(x)$ denotes the inverse function $\arcsin(x)$, whereas $(\sin(x))^{-1}$ denotes the reciprocal $\frac{1}{\sin(x)} = \csc(x)$.

What are the horizontal asymptotes of the graph of $y = \arctan(x)$?

Graph of y equals arctan(x) passing through the origin, bounded between dashed horizontal asymptotes at y equals pi over 2 and y equals negative pi over 2.

$$ y = \frac{\pi}{2} \quad \text{and} \quad y = -\frac{\pi}{2} $$ As $x \to \infty$, $\arctan(x) \to \frac{\pi}{2}$, and as $x \to -\infty$, $\arctan(x) \to -\frac{\pi}{2}$.

Why does the standard sine function $f(x) = \sin(x)$ fail to have an inverse function across its entire domain $(-\infty, \infty)$?

Graph of the sine wave intersected at multiple points by a horizontal line, demonstrating that the sine function fails the horizontal line test and is not one-to-one across its domain.

Because $f(x) = \sin(x)$ is periodic, it fails the horizontal line test. Multiple distinct input angles yield the same output value, making the function not one-to-one.

What standard restricted domain is chosen for $y = \sin(x)$ to make it invertible?

Graph of y equals sine of x highlighting the restricted domain interval from negative pi over 2 to pi over 2 with a bold blue curve, covering the full range from negative 1 to 1 to demonstrate invertibility.

$$ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$ This interval includes every value in the range $[-1, 1]$ exactly once while containing $0$.

In which quadrants on the unit circle do the principal output angles of $y = \arcsin(x)$ lie?

Unit circle diagram highlighting the right half in Quadrants I and IV, defining the range of arcsin(x) from negative pi over 2 to pi over 2.

Quadrant I for positive inputs $(0 \le x \le 1)$ and Quadrant IV as negative angles $\left[-\frac{\pi}{2}, 0\right)$ for negative inputs $(-1 \le x < 0)$.

Inverse Trig Foundations and Domain Restrictions

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