Inverse Trigonometric Functions: Definitions, Properties, and Applications unit 4.7

Inverse Trigonometric Functions

Introduction to Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine the angle that corresponds to a given trigonometric value. While the basic trigonometric functions (sine, cosine, tangent) take an angle as input and return a ratio, their inverses take a ratio and return an angle. These functions are essential for solving equations involving trigonometric expressions and for applications in geometry and calculus.

• Notation: The inverse of sine is written as $\sin^{-1} x$ or $\arcsin x$. Similarly, $\cos^{-1} x = \arccos x$ and $\tan^{-1} x = \arctan x$.

• Interpretation: For example, $\sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$ means the angle whose sine is $\frac{1}{2}$ is $30^\circ$.

• Domain Restrictions: Not all real numbers are valid inputs for inverse trigonometric functions. Each function has a restricted domain and range to ensure it is a function (passes the vertical line test).

Domains and Ranges of Inverse Trigonometric Functions

The following table summarizes the domains and ranges for the principal inverse trigonometric functions:

Function	Domain	Range
$y = \sin^{-1} x$	$-1 \leq x \leq 1$	$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
$y = \cos^{-1} x$	$-1 \leq x \leq 1$	$0 \leq y \leq \pi$
$y = \tan^{-1} x$	$-\infty < x < \infty$	$-\frac{\pi}{2} < y < \frac{\pi}{2}$

Additional info: The ranges are chosen so that each inverse function is single-valued (one-to-one correspondence).

Examples of Evaluating Inverse Trigonometric Functions

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

  • Find the angle $y$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\sin y = \frac{\sqrt{3}}{2}$.

  • Answer: $y = \frac{\pi}{3}$ (or $60^\circ$).

• Example 2: $\cos^{-1}\left(\frac{\sqrt{2}}{2}\right)$

  • Find the angle $y$ in $[0, \pi]$ such that $\cos y = \frac{\sqrt{2}}{2}$.

  • Answer: $y = \frac{\pi}{4}$ (or $45^\circ$).

• Example 3: $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$

  • Find the angle $y$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan y = \frac{\sqrt{3}}{3}$.

  • Answer: $y = \frac{\pi}{6}$ (or $30^\circ$).

Calculator Use and Domain Issues

• When using a calculator, ensure it is in the correct mode (degrees or radians) as required by the problem.

• Attempting to evaluate an inverse trigonometric function outside its domain will result in an error or "no solution." For example, $\sin^{-1}(1.2)$ is undefined because $1.2$ is not in $[-1, 1]$.

• Example: $\cos^{-1}(0.7) \approx 0.8$ radians (calculator in radian mode).

Inverse Secant, Cosecant, and Cotangent

The inverse secant, cosecant, and cotangent functions are less common and are typically evaluated by rewriting them in terms of sine, cosine, or tangent:

• $\sec^{-1} x = \cos^{-1}\left(\frac{1}{x}\right)$, where $|x| \geq 1$, $0 \leq y \leq \pi$, $y \neq \frac{\pi}{2}$

• $\csc^{-1} x = \sin^{-1}\left(\frac{1}{x}\right)$, where $|x| \geq 1$, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$, $y \neq 0$

• $\cot^{-1} x = \tan^{-1}\left(\frac{1}{x}\right)$, where $x \in \mathbb{R}$, $0 < y < \pi$

Additional info: These definitions allow for calculator evaluation and ensure the correct range for each function.

Examples:

• Find $\csc^{-1}\left(-2\right)$:

  • Rewrite as $\sin^{-1}\left(-\frac{1}{2}\right)$.

  • Find the angle $y$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin y = -\frac{1}{2}$.

  • Answer: $y = -\frac{\pi}{6}$.

• Find $\sec^{-1}(2)$:

  • Rewrite as $\cos^{-1}\left(\frac{1}{2}\right)$.

  • Find the angle $y$ in $[0, \pi]$ with $\cos y = \frac{1}{2}$.

  • Answer: $y = \frac{\pi}{3}$.

• Find $\cot^{-1}(-3)$:

  • Rewrite as $\tan^{-1}\left(-\frac{1}{3}\right)$.

  • Find the angle $y$ in $(0, \pi)$ with $\tan y = -\frac{1}{3}$ (second quadrant).

  • Answer: $y = \frac{5\pi}{6}$.

Cancellation Properties of Inverse Trigonometric Functions

Inverse trigonometric functions can "undo" their corresponding trigonometric functions, but only within the restricted domains and ranges:

• $\sin^{-1}(\sin x) = x$ if $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$

• $\cos^{-1}(\cos x) = x$ if $0 \leq x \leq \pi$

• $\tan^{-1}(\tan x) = x$ if $-\frac{\pi}{2} < x < \frac{\pi}{2}$

• For $\sin(\sin^{-1} x) = x$ if $-1 \leq x \leq 1$ (and similarly for cosine and tangent).

Additional info: If the input is outside the principal range, the result must be adjusted to fall within the correct interval.

Examples:

• Find $\tan(\tan^{-1}(3.5))$: Since $x$ can be any real number, the answer is $3.5$.

• Find $\sin^{-1}(\sin(\frac{3\pi}{2}))$: $\frac{3\pi}{2}$ is not in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so the answer is not $\frac{3\pi}{2}$. Instead, evaluate $\sin(\frac{3\pi}{2}) = -1$, so $\sin^{-1}(-1) = -\frac{\pi}{2}$.

Evaluating Compositions and Using Right Triangles

When evaluating expressions such as $\sin(\cos^{-1} x)$, it is helpful to use a right triangle to represent the relationships between the sides and the angle. This method is especially useful for algebraic expressions.

• Draw a right triangle where the angle $\theta = \cos^{-1} x$.

• Label the adjacent side as $x$ and the hypotenuse as $1$.

• Use the Pythagorean theorem to find the opposite side: $\sqrt{1 - x^2}$.

• Then $\sin(\cos^{-1} x) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}$.

Examples:

• Find $\sin(\cos^{-1}(\frac{3}{5}))$:

  • Draw a triangle with adjacent $3$, hypotenuse $5$.

  • Opposite side: $\sqrt{5^2 - 3^2} = 4$.

  • So $\sin(\cos^{-1}(\frac{3}{5})) = \frac{4}{5}$.

• Find $\cos(\sin^{-1}(\frac{-1}{6}))$:

  • Opposite side: $-1$, hypotenuse: $6$.

  • Adjacent side: $\sqrt{6^2 - (-1)^2} = \sqrt{35}$.

  • So $\cos(\sin^{-1}(\frac{-1}{6})) = \frac{\sqrt{35}}{6}$.

• Find $\tan(\cos^{-1}(\frac{-3}{4}))$:

  • Adjacent side: $-3$, hypotenuse: $4$.

  • Opposite side: $\sqrt{4^2 - (-3)^2} = \sqrt{7}$.

  • So $\tan(\cos^{-1}(\frac{-3}{4})) = \frac{\sqrt{7}}{-3}$.

• Find $\csc(\tan^{-1}(\frac{-1}{3}))$:

  • Opposite side: $-1$, adjacent: $3$.

  • Hypotenuse: $\sqrt{(-1)^2 + 3^2} = \sqrt{10}$.

  • So $\csc(\tan^{-1}(\frac{-1}{3})) = \frac{\sqrt{10}}{-1} = -\sqrt{10}$.

Algebraic Expressions with Inverse Trigonometric Functions

• Example: $\cos(\sin^{-1} u)$, $u > 0$

  • Draw a triangle with opposite $u$, hypotenuse $1$.

  • Adjacent side: $\sqrt{1 - u^2}$.

  • So $\cos(\sin^{-1} u) = \sqrt{1 - u^2}$.

• Example: $\sec(\tan^{-1} 4u)$, $u > 0$

  • Opposite side: $4u$, adjacent: $1$.

  • Hypotenuse: $\sqrt{1 + 16u^2}$.

  • So $\sec(\tan^{-1} 4u) = \frac{\sqrt{1 + 16u^2}}{1} = \sqrt{1 + 16u^2}$.

Summary Table: Inverse Trigonometric Functions

Function	Domain	Range	Alternate Notation
$\sin^{-1} x$	$-1 \leq x \leq 1$	$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$	$\arcsin x$
$\cos^{-1} x$	$-1 \leq x \leq 1$	$0 \leq y \leq \pi$	$\arccos x$
$\tan^{-1} x$	$-\infty < x < \infty$	$-\frac{\pi}{2} < y < \frac{\pi}{2}$	$\arctan x$

Additional info: Inverse trigonometric functions are essential for solving equations, evaluating integrals, and modeling periodic phenomena in advanced mathematics and applied sciences.