Let's find the exact values for each of the given expressions. These involve inverse trigonometric functions, so we need to determine angles based on known values of sine and cosine.

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

We are looking for the angle $\theta$ such that $\sin \theta = 1$. The sine of $\frac{\pi}{2}$ (90 degrees) is 1, and since the inverse sine function only takes values from $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the answer is:

$\sin^{-1}(1) = \frac{\pi}{2}$

b) $\cos^{-1} \left( \cos \frac{7\pi}{6} \right)$

The cosine of $\frac{7\pi}{6}$ is negative, as the angle is in the third quadrant. Specifically, $\cos \left( \frac{7\pi}{6} \right) = -\frac{\sqrt{3}}{2}$. The inverse cosine will return the principal angle between $0$ and $\pi$. The reference angle for $\frac{7\pi}{6}$ is $\frac{\pi}{6}$, so:

$\cos^{-1} \left( \cos \frac{7\pi}{6} \right) = \frac{7\pi}{6}$

c) $\sin^{-1} \frac{\sqrt{2}}{2}$

We need to find the angle $\theta$ such that $\sin \theta = \frac{\sqrt{2}}{2}$. This corresponds to an angle of $\frac{\pi}{4}$ (or 45 degrees), and since $\sin^{-1}$ returns values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we get:

$\sin^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}$

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

We need to find the angle $\theta$ such that $\cos \theta = -\frac{\sqrt{2}}{2}$. This corresponds to an angle of $\frac{3\pi}{4}$ (135 degrees), because $\cos$ is negative in the second quadrant. Thus:

$\cos^{-1} \left( -\frac{\sqrt{2}}{2} \right) = \frac{3\pi}{4}$

e) $\cos^{-1} 0$

We are looking for the angle $\theta$ such that $\cos \theta = 0$. This occurs at $\frac{\pi}{2}$ (90 degrees), so:

$\cos^{-1}(0) = \frac{\pi}{2}$

f) $\sin^{-1} \left( \sin \frac{7\pi}{6} \right)$

We know that $\sin \left( \frac{7\pi}{6} \right) = -\frac{1}{2}$. The inverse sine function will return the angle within $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ that gives $-\frac{1}{2}$. The corresponding angle is $-\frac{\pi}{6}$, so:

$\sin^{-1} \left( \sin \frac{7\pi}{6} \right) = -\frac{\pi}{6}$

Final Answers:

a) $\frac{\pi}{2}$
b) $\frac{7\pi}{6}$
c) $\frac{\pi}{4}$
d) $\frac{3\pi}{4}$
e) $\frac{\pi}{2}$
f) $-\frac{\pi}{6}$

Would you like further details on any of these calculations?

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Here are 5 related questions you might want to explore:

1. What is $\sin^{-1}(-1)$?
2. How do you find $\cos^{-1}(-1)$?
3. Can you explain why the domain of $\sin^{-1}(x)$ is restricted to $[-1, 1]$?
4. What is the range of $\cos^{-1}(x)$?
5. How do inverse trigonometric functions differ from regular trigonometric functions?

Tip: The range of inverse sine (arcsin) is always between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, so keep that in mind when calculating values!