To solve these, let's recall that the function $\sin^{-1}(x)$, or the inverse sine function (arcsin), is defined only for values $x$ in the range $[-1, 1]$. The output (or range) of $\sin^{-1}(x)$ is restricted to $[- \frac{\pi}{2}, \frac{\pi}{2}]$.

Solution

(a) $\sin^{-1}(-1)$

Since $-1$ is within the range $[-1, 1]$, we can evaluate it: $\sin^{-1}(-1) = -\frac{\pi}{2}$ This is because $\sin\left(-\frac{\pi}{2}\right) = -1$.

Answer: $-\frac{\pi}{2}$

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(b) $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$

Since $\frac{\sqrt{2}}{2}$ is within the range $[-1, 1]$, we can evaluate it: $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ This is because $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

Answer: $\frac{\pi}{4}$

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(c) $\sin^{-1}(-5)$

Since $-5$ is outside the range $[-1, 1]$, $\sin^{-1}(-5)$ is undefined.

Answer: UNDEFINED

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Would you like a more detailed explanation for any of these solutions?

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Related Questions

1. What is the exact value of $\sin^{-1}(1)$?
2. How would you evaluate $\cos^{-1}(-1)$?
3. Why is the domain of $\sin^{-1}(x)$ limited to $[-1, 1]$?
4. How do you evaluate $\sin\left(\sin^{-1}(x)\right)$ for $x$ within $[-1, 1]$?
5. What is the value of $\tan^{-1}(1)$ in terms of $\pi$?

Tip

When working with inverse trigonometric functions, always check if the input lies within the domain, as values outside of the defined range will be undefined.