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Chapter 4: Problem 100

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

Short Answer

Expert verified
The sine function must be restricted to the closed interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\) in order to have an inverse function.

Step by step solution

01

Understanding the sine function

The sine function is a periodic function with a period of \(2\pi\). It ranges from -1 to 1. The sine function is not one-to-one, it fails the horizontal line test, which states that any horizontal line will intersect the graph at most once to ensure that the inverse of a function will also be a function.
02

Restricting the sine function

In order to make the sine function one-to-one, and therefore invertible, we must restrict its domain. The common practice is to limit its domain to \([- \frac{\pi}{2}, \frac{\pi}{2}]\). In this restricted domain, the sine function is increasing and continuous, allowing it to pass the horizontal line test and have an inverse function.
03

Defining the inverse function

Once the sine function is restricted to \([- \frac{\pi}{2}, \frac{\pi}{2}]\), it becomes one-to-one, and its inverse function, denoted as \(\sin^{-1}(x)\) or \(\text{arcsin}(x)\), can be defined. The inverse sine function \(\sin^{-1}(x)\) takes a real number from -1 to 1 (the range of the sine function), and returns an angle in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

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Most popular questions from this chapter

Use the keys on your calculator or graphing utility for converting an angle in degrees, minutes, and seconds \(\left(D^{\circ} M^{\prime} S^{\prime \prime}\right)\) into decimal form, and vice versa. Convert each angle to \(D^{\circ} M^{\prime} S^{\prime \prime}\) form. Round your answer to the nearest second. $$50.42^{\circ}$$

will help you prepare for the material covered in the next section. a. Graph \(y=-3 \cos \frac{x}{2}\) for \(-\pi \leq x \leq 5 \pi\) b. Consider the reciprocal function of \(y=-3 \cos \frac{x}{2}\) namely, \(y=-3 \sec \frac{x}{2} .\) What does your graph from part (a) indicate about this reciprocal function for \(x=-\pi, \pi, 3 \pi,\) and \(5 \pi ?\)

Write the equation for a cosecant function satisfying the given conditions. $$\text { period: } 2 ; \text { range: }(-\infty,-\pi] \cup[\pi, \infty)$$

Use a graphing utility to graph each pair of functions in the same viewing rectangle. Use a viewing rectangle that shows the graphs for at least two periods. $$y=-3.5 \cos \left(\pi x-\frac{\pi}{6}\right) \text { and } y=-3.5 \sec \left(\pi x-\frac{\pi}{6}\right)$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Determine the range of each of the following functions. Then give a viewing rectangle, or window, that shows two periods of the function's graph. a. \(f(x)=3 \sin \left(x+\frac{\pi}{6}\right)-2\) b. \(g(x)=\sin 3\left(x+\frac{\pi}{6}\right)-2\)
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