Which value is similar to sin-1sin(6 π/7)?

The correct option is sin^-1(π/7).

Let's understand the step-by-step mathematical reasoning to show why this is the correct answer.

First, let's analyze the expression given in the question:
$${\sin }^{-1}\left(\sin \left(\frac{6\pi }{7}\right)\right)$$

We know that the principal value branch of the inverse sine function, ${\sin }^{-1}\left(x\right)$, is restricted to the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$.
This means that ${\sin }^{-1}\left(\sin \right(\theta \left)\right)=\theta$ only if $\theta$ lies within $[-\frac{\pi }{2},\frac{\pi }{2}]$.

Here, our angle is $\theta =\frac{6\pi }{7}$.
Since $\frac{6\pi }{7}>\frac{\pi }{2}$, the angle does not lie in the principal value branch. Therefore, we must use trigonometric identities to rewrite the angle within the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$.

We use the trigonometric identity $\sin (\pi -x)=\sin \left(x\right)$.
Let us express $\frac{6\pi }{7}$ in terms of $\pi$:
$$\frac{6\pi }{7}=\pi -\frac{\pi }{7}$$

Applying the identity:
$$\sin \left(\frac{6\pi }{7}\right)=\sin (\pi -\frac{\pi }{7})=\sin \left(\frac{\pi }{7}\right)$$

Substitute this back into our original expression:
$${\sin }^{-1}\left(\sin \left(\frac{6\pi }{7}\right)\right)={\sin }^{-1}\left(\sin \left(\frac{\pi }{7}\right)\right)$$

Since $\frac{\pi }{7}$ lies within the principal value interval $[-\frac{\pi }{2},\frac{\pi }{2}]$, we can simplify the expression:
$${\sin }^{-1}\left(\sin \left(\frac{\pi }{7}\right)\right)=\frac{\pi }{7}$$

Thus, ${\sin }^{-1}\sin \left(\frac{6\pi }{7}\right)$ is equivalent to ${\sin }^{-1}\left(\sin \right(\frac{\pi }{7}\left)\right)$, which corresponds exactly to the option sin^-1(π/7).