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Chapter 5: Problem 100

Describe the restriction on the sine function so that it has an inverse function.

Short Answer

Expert verified
To ensure the sine function has an inverse function, restrict the domain to the interval \([- \pi/2, \pi/2]\).

Step by step solution

01

Understanding Inverse Function

We need to find the condition under which the sine function \(y=\sin(x)\) is one-to-one. An inverse function exists when the original function is both one-to-one (there's only one output for every input) and onto (every possible output is paired with at least one input). Since sine function isn't one-to-one (For example, the \(\sin(0)\) and \(\sin(\pi )\) equal each other), it's necessary to restrict the domain.
02

Choosing the right interval for Sine function

To make sine function one-to-one, we pick the interval \([- \pi/2, \pi/2]\). This interval is the most commonly used because it keeps the sine function continuous and includes the maximum and minimum values of -1 and 1.
03

Confirming the existence of Inverse Function

With the restriction of \(x\) to the interval \([- \pi/2, \pi/2]\), the sine function is one-to-one and thus has an inverse function. The inverse of the sine function is denoted as arcsin(x) or \(\sin^{-1}(x)\).

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