Questions: Why do the functions f(x) = sin^(-1)(x) and g(x) = cos^(-1)(x) have different ranges?

Solution Steps

To determine why the functions \( f(x) = \sin^{-1}(x) \) and \( g(x) = \cos^{-1}(x) \) have different ranges, we need to consider the intervals where the sine and cosine functions are one-to-one. The arcsine function, \( \sin^{-1}(x) \), is defined for \( x \) in the interval \([-1, 1]\) and has a range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\). The arccosine function, \( \cos^{-1}(x) \), is also defined for \( x \) in the interval \([-1, 1]\) but has a range of \([0, \pi]\). These ranges are different because the largest intervals where \(\sin(x)\) and \(\cos(x)\) are one-to-one do not coincide.

The function \( f(x) = \sin^{-1}(x) \) is defined for \( x \in [-1, 1] \). The range of this function is given by: \[ \text{Range of } f(x) = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \approx (-1.5708, 1.5708) \]

The function \( g(x) = \cos^{-1}(x) \) is also defined for \( x \in [-1, 1] \). The range of this function is: \[ \text{Range of } g(x) = [0, \pi] \approx (0, 3.1416) \]

The ranges of the two functions are:

• \( f(x) = \sin^{-1}(x) \) has a range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
• \( g(x) = \cos^{-1}(x) \) has a range of \( [0, \pi] \)

Since these ranges do not overlap, we conclude that the functions \( f(x) \) and \( g(x) \) have different ranges.

Final Answer