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?
 A: 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.

A: An intuitive way to see this is to get that sin is a monotonically increasing function from 0 to 90 degrees.So in this range that sinr-sini=1/2(sin i) or sinr-sini vs sini has a constant +ve slope.Now the nature of graph remains same if instead of a monotonous function we have only their variable.So the graph of r-i vs i is a straight line with constant +ve slope.But for how long? Obviously as long as 3/2 sin i< 1 or sin i<2/3 the value of i is close to 45 but lesser than that.
A: Clearly, the "straight" solution is
$$
{3 \over 2}\sin i = \sin r\quad  \Leftrightarrow \quad r = \arcsin \left( {{3 \over 2}\sin i} \right)\quad  \Leftrightarrow \quad i = \arcsin \left( {{2 \over 3}\sin r} \right)
$$
But, just to catch the behaviour of the solution at lower value of the angles, we can use the Taylor series for $\sin$ to get
$$
\eqalign{
  & {3 \over 2}\sin i = \sin r  \cr 
  & {3 \over 2}\left( {i - {{i^3 } \over 6} + O\left( {i^5 } \right)} \right) = r - {{r^3 } \over 6} + O\left( {r^5 } \right)  \cr 
  & {3 \over 2}i\left( {1 - {{i^2 } \over 6}} \right) \cong r\left( {1 - {{r^2 } \over 6}} \right)  \cr 
  & {3 \over 2}i \cong r \cr} 
$$
Of course you can increase the precision of the approximation
by solving the cubic , instead of the last linear approximation.
Otherwise you can use the Taylor series for $\arcsin$, etc.
