Let's evaluate the inverse trigonometric functions given in the question and express the answers in radians and in terms of $\pi$.

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a) $\sin^{-1}\left(\frac{3}{2}\right)$

• The sine function only outputs values in the range $[-1, 1]$. Since $\frac{3}{2}$ is greater than 1, this expression is undefined.

$\boxed{\text{Undefined}}$

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

• $\frac{2}{2} = 1$.
• The cosine of 0 radians (or 0 degrees) is 1. Therefore:

$\cos^{-1}(1) = 0$

So,

$\boxed{0}$

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c) $\tan^{-1}(1)$

• The tangent of $\frac{\pi}{4}$ radians (or 45 degrees) is 1. Therefore:

$\tan^{-1}(1) = \frac{\pi}{4}$

So,

$\boxed{\frac{\pi}{4}}$

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d) $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

• The sine of $-\frac{\pi}{3}$ radians (or $-60$ degrees) is $-\frac{\sqrt{3}}{2}$. Therefore:

$\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$

So,

$\boxed{-\frac{\pi}{3}}$

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e) $\cos^{-1}(-1)$

• The cosine of $\pi$ radians (or 180 degrees) is $-1$. Therefore:

$\cos^{-1}(-1) = \pi$

So,

$\boxed{\pi}$

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f) $\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)$

• The tangent of $-\frac{\pi}{6}$ radians (or $-30$ degrees) is $-\frac{\sqrt{3}}{3}$. Therefore:

$\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) = -\frac{\pi}{6}$

So,

$\boxed{-\frac{\pi}{6}}$

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Do you have any questions or would you like to go over the details of any of these solutions?

Here are five related questions you might find helpful:

1. What is $\sin^{-1}(0)$?
2. What is the range of the $\cos^{-1}(x)$ function?
3. How do you evaluate $\tan^{-1}(\infty)$?
4. What is the relationship between $\sin^{-1}(x)$ and $\cos^{-1}(x)$?
5. Can you calculate $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$?

Tip: Remember that the domain and range of inverse trigonometric functions are key to determining valid outputs.