Identity involving $\arcsin$ and $\sin$. $$\sin z = \sin \left(\arcsin \frac{\pi}{6} - \arccos \frac{\pi}{6} \right)$$
Is there a specific expansion of the expression on the right hand side of the equation? To evaluate the value of $z$? 

As pointed out by @Bye_World, this looks like the trigonometric identity:
$$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \tag{1}$$
Since you already have:
$$arcsin \frac{\pi}{6} \Rightarrow \sin \alpha = \frac{\pi}{6} \tag{2}$$
$$arccos \frac{\pi}{6} \Rightarrow \cos \beta = \frac{\pi}{6} \tag{3}$$
You can then calculate the opposite sides:
$$\cos \alpha = \sqrt{1 - (\frac{\pi}{6})^2} \tag{4}$$
$$\sin \beta = \sqrt{1 - (\frac{\pi}{6})^2} \tag{5}$$
If you replace $(2, 3, 4, 5)$ in $(1)$, you now have:
$$\sin(\alpha - \beta) = \frac{\pi^2}{18} - 1 \tag{6}$$
The value of $z$ is therefore the $\arcsin$ of $(6)$.
