To evaluate $\sin^{-1}(\sin(7\pi/6))$, let’s carefully consider the behavior of the inverse sine function $\sin^{-1}$ (also known as arcsin):

Steps:

1. Simplify $\sin(7\pi/6)$:

   • $\sin(7\pi/6)$ corresponds to an angle in the third quadrant where sine is negative.
   • The reference angle for $7\pi/6$ is $\pi/6$, and since it is in the third quadrant, $\sin(7\pi/6) = -\sin(\pi/6) = -\frac{1}{2}$.

   Therefore, $\sin(7\pi/6) = -\frac{1}{2}$.

2. Find $\sin^{-1}(-\frac{1}{2})$:

   • The function $\sin^{-1}$ (arcsin) returns values in the range $[-\pi/2, \pi/2]$.
   • For $\sin^{-1}(-\frac{1}{2})$, the angle within this range that satisfies $\sin(\theta) = -\frac{1}{2}$ is $\theta = -\pi/6$.

Final Answer:

$\sin^{-1}(\sin(7\pi/6)) = -\frac{\pi}{6}.$

Would you like a detailed explanation of inverse trigonometric properties or additional examples?

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Follow-up Questions:

1. What are the domain and range of $\sin^{-1}(x)$?
2. How can you verify $\sin^{-1}(\sin(x))$ for any angle $x$?
3. What happens if the input angle is not in the principal range of arcsine?
4. Can you generalize the result for $\sin^{-1}(\sin(n\pi + x))$?
5. How does $\cos^{-1}(\cos(x))$ differ in behavior from $\sin^{-1}(\sin(x))$?

Tip: Always remember that the range of $\sin^{-1}(x)$ is restricted to $[-\pi/2, \pi/2]$, which is crucial for interpreting results.