$$ \sin^{-1}^{\placeholder{}} $$(sin$$ \frac17 $$$$ \pi5) $$

It looks like you are working with an expression involving inverse trigonometric functions. The expression you wrote seems to be:

\[
\sin^{-1}(\sin(\frac{5\pi}{7}))
\]

To solve this, we need to evaluate the arcsine of $\sin(\frac{5\pi}{7})$.

### Key points:
1. The sine function is periodic with a period of $2\pi$.
2. The range of the arcsine function, $\sin^{-1}(x)$, is $[- \frac{\pi}{2}, \frac{\pi}{2}]$.
3. The value $\frac{5\pi}{7}$ is between 0 and $\pi$, so we need to adjust it to fit within the principal range of $\sin^{-1}$.

We know that:
\[
\sin(\frac{5\pi}{7}) = \sin(\pi - \frac{5\pi}{7}) = \sin(\frac{2\pi}{7})
\]
Therefore:
\[
\sin^{-1}(\sin(\frac{5\pi}{7})) = \frac{2\pi}{7}
\]

Thus, the value of the given expression is:

\[
\sin^{-1}(\sin(\frac{5\pi}{7})) = \frac{2\pi}{7}
\]

Would you like more details on how the sine and inverse sine functions interact?

### Related questions:
1. What is the general range of the arcsine function?
2. How do the sine and arcsine functions cancel each other under specific conditions?
3. What is the sine of angles larger than $\pi$?
4. How can we extend this idea to other inverse trigonometric functions?
5. Can we generalize this process for values of sine outside $[-\pi, \pi]$?

**Tip:** Remember, the arcsine function only returns values within the restricted range $[- \frac{\pi}{2}, \frac{\pi}{2}]$, which means for angles outside this range, sine must be adjusted accordingly.

$$ \sin^{-1}^{\placeholder{}} $$(sin$$ \frac{5\pi}{7} $$)

Discover how to solve the trigonometric expression \sin^{-1}(\sin(\frac{5\pi}{7})). This solution covers key trigonometry concepts such as periodicity of the sine function and the range of the arcsine function. Suitable for high school students in Grades 10-12.

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