To find the exact value of $\sin^{-1}(\sin(11\pi/12))$, we need to consider the properties of the inverse sine ($\sin^{-1}$) function. Here's how to compute it:

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Step-by-Step Solution:

1. Domain and Range of $\sin^{-1}$: The inverse sine function, $\sin^{-1}(x)$, outputs values in the range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$ So, the result of $\sin^{-1}(\sin(\theta))$ must lie within this range.

2. Simplify the Argument $11\pi/12$: The given angle $11\pi/12$ is greater than $\pi/2$ (since $\pi/2 = 6\pi/12$) and less than $\pi$ (since $\pi = 12\pi/12$). This places it in the second quadrant.

3. Reference Angle for $11\pi/12$: In the second quadrant, the sine value is positive, and the reference angle is: $\text{Reference Angle} = \pi - \theta = \pi - \frac{11\pi}{12} = \frac{\pi}{12}$

4. Adjust for the Range of $\sin^{-1}$: To bring the angle into the range of $[- \frac{\pi}{2}, \frac{\pi}{2}]$, we need an equivalent angle with the same sine value. For $11\pi/12$ in the second quadrant, the equivalent angle is: $\sin^{-1}(\sin(11\pi/12)) = \frac{\pi}{12}$

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Final Answer:

$\sin^{-1}(\sin(11\pi/12)) = \frac{\pi}{12}$

Would you like further clarification or a visual explanation? 😊

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5 Related Questions:

1. What is the general formula for $\sin^{-1}(\sin(\theta))$ when $\theta$ is outside the principal range?
2. How does the symmetry of the sine function affect the calculation of inverse sine?
3. What is the exact value of $\sin^{-1}(\sin(7\pi/6))$?
4. How can reference angles be used to simplify trigonometric problems?
5. Why is the range of $\sin^{-1}(x)$ restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$?

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Tip:

Always consider the principal range of inverse trigonometric functions when solving problems to ensure your answer is valid.