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3.9 Inverse Trigonometric Functions in AP Precalculus

3.9 Inverse Trigonometric Functions in AP Precalculus
Table of Contents

1. Introduction

Inverse trigonometric functions play a crucial role in AP Precalculus, helping us reverse the process of standard trigonometric functions. While functions like sine, cosine, and tangent take an angle as input and return a ratio, their inverses work in the opposite direction—starting with a ratio and outputting an angle.

Understanding inverse trigonometric functions is essential for solving real-world problems involving angles and distances, such as physics applications, engineering problems, and even computer graphics. There are many more interesting applications which can be found in the resources we enlisted in Top 10 books for AP Precalculus. However, because trigonometric functions are not naturally one-to-one, their domains must be restricted to ensure each inverse function is well-defined.

In this blog, we will explore the domain and range of inverse trigonometric functions, analyze their graphs, understand the significance of principal values, and go through solved examples that demonstrate how to find inverses, determine valid domains, and graph inverse trig functions. By the end, you’ll have a strong grasp of these functions and be better prepared for AP Precalculus problems. Let’s dive in!

2. Inverse Trigonometric Functions

Inverse trigonometric functions are the mathematical tools that allow us to reverse the effect of standard trigonometric functions. In other words, if a trigonometric function like sin(θ) = x gives us a ratio when we input an angle, the inverse function sin⁻¹(x) (also called arcsin) helps us determine the angle when given the ratio.

For example:

\sin(30^\circ) = \frac{1}{2} \quad \Rightarrow \quad \sin^{-1}(\frac{1}{2}) = 30^\circ

This concept applies similarly to other trigonometric functions like cosine, tangent, cotangent, secant, and cosecant.

However, unlike basic algebraic functions, trigonometric functions are periodic, meaning they repeat their values infinitely. This creates a problem when defining their inverses because a function must be one-to-one to have a proper inverse. To solve this, we restrict the domain of trigonometric functions so that each inverse function is uniquely defined.

Notation for Inverse Trigonometric Functions

  • \sin^{-1}(x) or arcsin(x)
  • \cos^{-1}(x) or arccos(x)
  • \tan^{-1}(x) or arctan(x)
  • \cot^{-1}(x) or arccot(x)
  • \sec^{-1}(x) or arcsec(x)
  • \csc^{-1}(x) or arccsc(x)

3. Domain and Range of Inverse Trigonometric Functions

To properly define inverse trigonometric functions, we need to restrict the domain of their corresponding trigonometric functions so that they become one-to-one and pass the horizontal line test. This ensures that each input has only one unique output, making the inverse function well-defined.

Domain and Range of Basic Inverse Trigonometric Functions

FunctionDomain (Input Values)Range (Output Values)
\sin^{-1}(x) (arcsin)[-1,1]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
\cos^{-1}(x) (arccos)[-1,1][0, \pi]
\tan^{-1}(x) (arctan)(-\infty, \infty)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
\cot^{-1}(x) (arccot)(-\infty, \infty)(0, \pi)
\sec^{-1}(x) (arcsec)(-\infty, -1] \cup [1, \infty)[0, \pi] \quad x \neq \frac{\pi}{2}
\csc^{-1}(x) (arccsc)(-\infty, -1] \cup [1, \infty)\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \quad y \neq 0

Key Observations

  1. Inverse sine and cosine are only defined for inputs between -1 and 1, because the sine and cosine functions themselves never produce values outside this range.
  2. Inverse tangent and cotangent are defined for all real numbers, but their range is limited to keep them one-to-one.
  3. Inverse secant and cosecant require inputs greater than or equal to 1 or less than or equal to -1, since the original sec and csc functions never output values between -1 and 1.

Example 1: Determining the Domain and Range

Find the domain and range of f(x) = \sin^{-1} (2x - 1).

Solution:

  1. Since the domain of \sin^{-1}x is [-1,1], we set: -1 \leq 2x - 1 \leq 1
  2. Solving for x: 0 \leq 2x \leq 2 0 \leq x \leq 1 So, the domain is [0,1].
  3. The range of \sin^{-1} (2x - 1) is the same as the basic \sin^{-1}x function, which is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Download this free AP Precalculus worksheet with solutions for targeted practice and a better understanding of concepts

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4. Graphs of Inverse Trigonometric Functions

Understanding the graphs of inverse trigonometric functions is essential for visualizing their behavior, identifying their domain and range, and solving problems efficiently. Since inverse functions swap the roles of inputs and outputs, their graphs can be obtained by reflecting the original trigonometric function across the line y = x.

Graphs of Basic Inverse Trigonometric Functions

  1. Graph of \sin^{-1}(x) (Arcsin)

    • Domain: [-1,1]
    • Range: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
    • Key Points:
      • \sin^{-1}(-1) = -\frac{\pi}{2}
      • \sin^{-1}(0) = 0
      • \sin^{-1}(1) = \frac{\pi}{2}

3.9 Inverse Trigonometric Functions in AP Precalculus

  1. Graph of \cos^{-1}(x) (Arccos)
    • Domain: [-1,1]
    • Range: <span class="katex">[0, \pi]
    • Key Points:
      • \cos^{-1}(-1) = \pi
      • \cos^{-1}(0) = \frac{\pi}{2}
      • \cos^{-1}(1) = 0

3.9 Inverse Trigonometric Functions in AP Precalculus

  1. Graph of \tan^{-1}(x) (Arctan)
    • Domain: (-\infty, \infty)
    • Range: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
    • Asymptotes: y = \pm \frac{\pi}{2}
    • Key Points:
      • \tan^{-1}(-\infty) = -\frac{\pi}{2}
      • \tan^{-1}(0) = 0
      • \tan^{-1}(\infty) = \frac{\pi}{2}

3.9 Inverse Trigonometric Functions in AP Precalculus

  1. Graph of \cot^{-1}(x) (Arccot)
    • Domain: (-\infty, \infty)
    • Range: (0, \pi)
    • Asymptotes: y = 0 and y = \pi

3.9 Inverse Trigonometric Functions in AP Precalculus

  1. Graph of \sec^{-1}(x) (Arcsec)
    • Domain: (-\infty, -1] \cup [1, \infty)
    • Range: [0, \pi], y \neq \frac{\pi}{2}

3.9 Inverse Trigonometric Functions in AP Precalculus

  1. Graph of csc⁡−1(x)\csc^{-1}(x) (Arccsc)
    • Domain: (-\infty, -1] \cup [1, \infty)
    • Range: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], y \neq 0

3.9 Inverse Trigonometric Functions in AP Precalculus

Example: Graphing an Inverse Trigonometric Function

Graph f(x) = \sin^{-1}(2x - 1) for x \in [0,1].

Solution:

  1. Find the domain:
    • Since \sin^{-1} is only defined for values between [-1,1], we set: -1 \leq 2x - 1 \leq 1
    • Solving for x: 0 \leq x \leq 1
    • So, the function is defined for x \in [0,1].
  2. Find key points for graphing:
    • When x = 0, y = \sin^{-1}(-1) = -\frac{\pi}{2}.
    • When x = \frac{1}{2}, y = \sin^{-1}(0) = 0.
    • When x = 1, y = \sin^{-1}(1) = \frac{\pi}{2}.
  3. Graph the function (Attach the graph here).

By using reflections and transformations, we can easily sketch inverse trigonometric functions.

5. Principal Values of Inverse Trigonometric Functions

Inverse trigonometric functions must be one-to-one to have a well-defined inverse. However, trigonometric functions are periodic, meaning they repeat infinitely. To ensure each inverse function produces a unique output, we restrict the range to specific intervals called principal values.

Principal Value Ranges of Inverse Trigonometric Functions

Each inverse function has a unique principal value range, ensuring that the output remains consistent across all inputs.

FunctionPrincipal Value Range
\sin^{-1}(x)\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]
\cos^{-1}(x)[0, \pi]
\tan^{-1}(x)\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)
\cot^{-1}(x)(0, \pi)
\sec^{-1}(x)[0, \pi], \, y \neq \frac{\pi}{2}
\csc^{-1}(x)\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], \, y \neq 0

These ranges are chosen so that each inverse function produces only one valid answer for a given input.

Understanding Principal Values with Examples

✅ Example 1: Finding the Principal Value of \sin^{-1}(-\frac{1}{2})
We need to find an angle \theta such that:

\sin \theta = -\frac{1}{2}

From the unit circle, \sin(-\frac{\pi}{6}) = -\frac{1}{2}. Since the principal range of arcsin is \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], we take:

\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}

✅ Example 2: Finding the Principal Value of \tan^{-1}(\sqrt{3})
We need to find an angle \theta such that:

\tan \theta = \sqrt{3}

Since \tan(\frac{\pi}{3}) = \sqrt{3} and the principal range of arctan is \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), we get:

\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

✅ Example 3: Finding the Principal Value of \cos^{-1}(-\frac{1}{2})
We need to find an angle \theta such that:

\cos \theta = -\frac{1}{2}

From the unit circle, \cos(\frac{2\pi}{3}) = -\frac{1}{2}. Since the principal range of arccos is [0, \pi], we take:

\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}

Some Problems3.9 Inverse Trigonometric Functions in AP Precalculus

Solution :

  • The point R(x, y) is on the unit circle where the terminal ray of <span class="katex">\theta intersects.
  • The points P, Q, and S are reflections:
    • P is the reflection over the y-axis, so its coordinates are (-x, y).
    • Q is the reflection over the origin, so its coordinates are (-x, -y).
    • S is the reflection over the x-axis, so its coordinates are (x, -y).

Now, we solve each problem statement.

Solution for each part:

(a) \sin^{-1}(y)

  • The function \sin^{-1}(y) gives an angle whose sine is y.
  • Looking at the unit circle, both R(x, y) and P(-x, y) share the same y-coordinate.
  • However, the principal value range for \sin^{-1}(y) is [-\frac{\pi}{2}, \frac{\pi}{2}] (quadrants I and IV).
  • Since R(x, y) is in quadrant I, R is the correct point.

✅ Answer: R(x, y)

(b) \cos^{-1}(y)

  • The function \cos^{-1}(y) gives an angle whose cosine is y.
  • Looking at the unit circle, both R(x, y) and S(x, -y) share the same x-coordinate.
  • The principal range for \cos^{-1}(x) is [0, \pi] (quadrants I and II).
  • Since R(x, y) is in quadrant I, R is the correct point.

✅ Answer: R(x, y)

(c) \tan^{-1}(\frac{y}{x})

  • The function \tan^{-1}(\frac{y}{x}) gives an angle whose tangent is \frac{y}{x}.
  • Looking at the unit circle, \tan \theta = \frac{y}{x} applies to point R(x, y).
  • The principal value range for \tan^{-1}(x) is (-\frac{\pi}{2}, \frac{\pi}{2}) (quadrants I and IV).
  • Since R(x, y) is in quadrant I, R is the correct point.

✅ Answer: R(x, y)

(d)\sin^{-1}(-y)

  • The function \sin^{-1}(-y) gives an angle whose sine is -y.
  • Looking at the unit circle, both S(x, -y) and Q(-x, -y) share the same y-coordinate.
  • The principal value range for \sin^{-1}(x) is [-\frac{\pi}{2}, \frac{\pi}{2}] (quadrants I and IV).
  • Since S(x, -y) is in quadrant IV, S is the correct point.

✅ Answer: S(x, -y)

(e) \cos^{-1}(-y)

  • The function \cos^{-1}(-x) gives an angle whose cosine is -x.
  • Looking at the unit circle, both P(-x, y) and Q(-x, -y) share the same x-coordinate.
  • The principal value range for \cos^{-1}(x) is [0, \pi] (quadrants I and II).
  • Since P(-x, y) is in quadrant II, P is the correct point.

✅ Answer: P(-x, y)

(f) \tan^{-1}(-\frac{y}{x})

  • The function \tan^{-1}(-\frac{y}{x}) gives an angle whose tangent is -\frac{y}{x}.
  • Looking at the unit circle, both S(x, -y) and P(-x, y) satisfy this condition.
  • The principal value range for \tan^{-1}(x) is (-\frac{\pi}{2}, \frac{\pi}{2}) (quadrants I and IV).
  • Since S(x, -y) is in quadrant IV, S is the correct point.

✅ Answer: S(x, -y)

Final Answers:

ExpressionPoint
\sin^{-1}(y)R(x, y)
\cos^{-1}(x)R(x, y)
\tan^{-1}(\frac{y}{x})R(x, y)
\sin^{-1}(-y)S(x, -y)
\cos^{-1}(-x)P(-x, y)
\tan^{-1}(-\frac{y}{x})S(x, -y)

This step-by-step breakdown ensures that each inverse trigonometric function is correctly matched with its corresponding reflected point on the unit circle.

FAQs : Frequently Asked Questions

 

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About Colin Phillips

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