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Chapter 7: Problem 34

State the ranges of the inverse sine, inverse cosine, and inverse tangent functions.

Short Answer

Expert verified
The range of \(\text{sin}^{-1}(x)\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), \(\text{cos}^{-1}(x)\) is \([0, \pi]\), and \(\text{tan}^{-1}(x)\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Step by step solution

01

Understanding Inverse Sine Function

The inverse sine function, denoted as \(\text{sin}^{-1}(x)\) or \(\text{arcsin}(x)\), takes a value from \([-1, 1]\) and returns the angle whose sine is that value. The range of \(\text{sin}^{-1}(x)\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
02

Understanding Inverse Cosine Function

The inverse cosine function, denoted as \(\text{cos}^{-1}(x)\) or \(\text{arccos}(x)\), takes a value from \([-1, 1]\) and returns the angle whose cosine is that value. The range of \(\text{cos}^{-1}(x)\) is \([0, \pi]\).
03

Understanding Inverse Tangent Function

The inverse tangent function, denoted as \(\text{tan}^{-1}(x)\) or \(\text{arctan}(x)\), takes a value from \((-\infty, \infty)\) and returns the angle whose tangent is that value. The range of \(\text{tan}^{-1}(x)\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

inverse sine function
The inverse sine function is a crucial concept in trigonometry. It is denoted as \(\text{sin}^{-1}(x)\) or \(\text{arcsin}(x)\). This function is used to find the angle whose sine value is a given number. For example, if you know the sine of an angle is 0.5, \(\text{sin}^{-1}(0.5)\) will tell you the angle that has this sine value.

Important points about the inverse sine function:
  • The domain (input values) is \([-1, 1]\).
  • The range (possible output angles) is \([-\frac{\text{蟺}}{2}, \frac{\text{蟺}}{2}]\).
  • This function is commonly used in various practical applications like calculating the height of a tree using angles or understanding wave functions in physics.

To summarize, the inverse sine function helps you map a sine value to its corresponding angle, making it an invaluable tool for solving trigonometric problems.
inverse cosine function
The inverse cosine function, expressed as \(\text{cos}^{-1}(x)\) or \(\text{arccos}(x)\), is another integral part of trigonometry. It tells you the angle whose cosine is equal to a given number. For instance, if the cosine of an angle is 0.5, \(\text{cos}^{-1}(0.5)\) will provide the angle with this cosine value.

Key points to remember:
  • The domain (input values) ranges from \([-1, 1]\).
  • The range (allowed output angles) is \([0, \text{蟺}]\).
  • This function is particularly useful in fields like engineering and architecture where precise angle measurements are essential.

In conclusion, the inverse cosine function allows you to find the original angle from a known cosine value, providing crucial information for solving various trigonometric and real-world problems.
inverse tangent function
The inverse tangent function is represented as \(\text{tan}^{-1}(x)\) or \(\text{arctan}(x)\). This function is used to determine the angle whose tangent is a specified value. For example, if the tangent of an angle is 1, \(\text{tan}^{-1}(1)\) will give you the angle that has this tangent value.

Key characteristics of the inverse tangent function:
  • The domain (input values) extends from \((-\text{鈭瀩, \text{鈭瀩)\), meaning it can handle all real numbers.
  • The range (output angles) is \((-\frac{\text{蟺}}{2}, \frac{\text{蟺}}{2})\).
  • This function is extremely useful in navigation, where determining angles and bearings is essential.

Thus, the inverse tangent function helps you uncover the original angle from a known tangent value, making it invaluable for both theoretical and practical applications.

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Most popular questions from this chapter

Solve, finding all solutions in \([0,2 \pi)\). $$\cos (\pi-x)+\sin \left(x-\frac{\pi}{2}\right)=1$$

Find the following. \(\cos \left(\frac{1}{2} \sin ^{-1} \frac{\sqrt{3}}{2}\right)\)

Solve in \([0,2 \pi)\),. $$|\sin x|=\frac{\sqrt{3}}{2}$$

Assuming that \(\sin u=\frac{3}{5}\) and \(\sin v=\frac{4}{5}\) and that \(u\) and \(v\) are between 0 and \(\pi / 2,\) evaluate each of the following exactly. $$\cos (u-v)$$

Solve in \([0,2 \pi)\),. $$\ln (\cos x)=0$$
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