Inverse trigonometric functions used for difference in angle? I was actually solving a physics question in which I got the equation 3/2 (sin i) = sin r. The graph of (r-i) against i has a constant positive slope till 45 degrees value for i. How do I get this result? The only thing I can get from the equation is that sin r - sin i = sin i/2. I don't have much of knowledge about inverse trigonometric functions, would someone please help me understand this in a simple way?

We should understand refraction optics through Snell's Law. From the figure below we can see that after $i> 90^{\circ}$there is a grey area where refracted ray cannot enter.
$$ \frac{\sin i}{\sin  r} = \mu = \frac{3}{2}$$
$$ \frac {\sin 90^{\circ}}  {\sin{r_{critical}}} = \mu = \frac{3}{2}$$
The critical limit angle limit or cut-off angle is $r_{crit}= 41.81$ degrees by calculation. 
That means in the reversed arrow sense if a fish is hypothetically  swimming in a covered liquid of this $\mu$ in the grey region outside the cone, it cannot see anything outside through a small hole at this incident point.
Another phenomenon total internal reflection occurs for higher angles of incidence.

