The given expression is:

$\sin^{-1}(\sin\left(\frac{7\pi}{4}\right)).$

To solve this:

1. First, simplify $\sin\left(\frac{7\pi}{4}\right)$.

   • The angle $\frac{7\pi}{4}$ is in the fourth quadrant, where the sine function is negative.
   • Using periodic properties of the sine function, we have: $\sin\left(\frac{7\pi}{4}\right) = \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}.$

2. Now, we need to find $\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.

   • The inverse sine function, $\sin^{-1}(x)$, returns values in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$.
   • Therefore, $\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$.

Thus, the final result is:

$\sin^{-1}(\sin\left(\frac{7\pi}{4}\right)) = -\frac{\pi}{4}.$

Would you like further details on any step? Here are some related questions you may find useful:

1. How does the sine function behave in different quadrants?
2. What are the periodic properties of trigonometric functions?
3. How do we calculate the principal value of inverse trigonometric functions?
4. Why is the range of the inverse sine function restricted to $[- \frac{\pi}{2}, \frac{\pi}{2}]$?
5. How can we simplify expressions involving trigonometric and inverse trigonometric functions?

Tip: Remember, the principal value of an inverse trigonometric function is always within a specified range. For $\sin^{-1}(x)$, it lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.