Problem 100 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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 \(\sin^{-1}(x)\).