To solve $\sin^{-1}(\sin(6\pi/7))$, we need to keep in mind that the inverse sine function, $\sin^{-1}(x)$, gives us an angle in the principal range $[- \frac{\pi}{2}, \frac{\pi}{2}]$, also called the principal value of the inverse sine.

Step-by-step solution:

1. Simplify the sine function: $\sin\left(\frac{6\pi}{7}\right)$ The angle $\frac{6\pi}{7}$ is between $\pi/2$ and $\pi$, which places it in the second quadrant. In the second quadrant, $\sin(\theta)$ is positive, and for $\theta = \frac{6\pi}{7}$, the sine is equivalent to the sine of its reference angle: $\sin\left(\frac{6\pi}{7}\right) = \sin\left(\pi - \frac{6\pi}{7}\right) = \sin\left(\frac{\pi}{7}\right)$

2. Apply the inverse sine: Now, we apply the inverse sine function, keeping in mind that $\sin^{-1}(x)$ returns an angle in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$. The sine of $\frac{6\pi}{7}$ is positive, but $\frac{6\pi}{7}$ is outside the range of the inverse sine function.

3. Find the principal value: To bring the angle into the correct range for the inverse sine function, we use the reference angle: $\sin^{-1}(\sin(6\pi/7)) = \sin^{-1}(\sin(\pi/7)) = \frac{\pi}{7}$ This is because $\pi/7$ lies within the valid range of the inverse sine function.

Final Answer:

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

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Would you like further details or have any questions?

Here are five related questions to explore:

1. What is the principal value of $\sin^{-1}(\sin(5\pi/4))$?
2. How do we handle inverse trigonometric functions for angles outside their principal ranges?
3. Can you explain how reference angles work in different quadrants?
4. What is the difference between $\sin^{-1}(x)$ and $\frac{1}{\sin(x)}$?
5. How does the periodicity of sine affect the calculation of inverse functions?

Tip: Always check the range of the inverse trigonometric function when solving problems like these, as they restrict the output angle to specific intervals.