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