In this section, we will be investigating the inverse trigonometric functions.
In practical terms, the inverse trigonometric functions are intended to answer the question of: if someone has a specific Sine, Cosine, or Tangent value, what is the corresponding angle that causes that value?
Consider this question (answer using the Unit Circle):
What angle results in \(\sin(\theta) = \frac{{1}}{{2}}\)?
Put another way, what is \(\theta\)?
Once we know the answer to this, we would be able to state that the answer is the inverse, or rather:
\(\theta = \sin^{-1}\left(\frac{{1}}{{2}}\right)\).
However, as you might have already noticed from the Unit Circle, there is a bit of a problem:
Which angle is it? In fact, while only two are drawn, there are infinitely many angles which result in the Sine function output of 1/2. Consider the graph of Sine:
As you can see, with a domain of all real numbers and by panning around, the output of 1/2 occurs all over the place. Our goal, then, must be to restrict the domain of the Sine function so that there is only one output for any given input.
In order to make our domain restriction sensible, we should also try to restrict the domain to as close to 0 as possible (basically try to fit it into the \([0,2\pi)\) if we can):
Unfortunately, as you can see, \([0,2\pi)\) doesn't work; we need it to be monotonic, either only increasing or only decreasing, on the whole domain restriction. Let's try \([-\pi,\pi]\) instead:
This is better, but we still have ups and downs. The trick, then, is to go from a minimum directly to a maximum. This happens on \(\left[-\frac{\pi}{{2}},\frac{\pi}{{2}}\right]\):
With the Domain restricted to \(\left[-\frac{\pi}{{2}},\frac{\pi}{{2}}\right]\), we now have a function that is invertible. Thus, whenever we use the inverse trig function, \(\sin^{-1}(x)\), it must have a domain of \([-1,1]\) with a range of \(\left[-\frac{\pi}{{2}},\frac{\pi}{{2}}\right]\) (why? remember, when we look at the inverse functions we flip the domain/range!)
Coming back to the Unit Circle, by limiting the domain of the Sine function as we have, we are effectively only looking at half the Unit Circle in order to guarantee that any horizontal line only intersects the Unit Circle once:
Finally, we can answer our original question in full:
\[ \theta = \sin^{-1}\left(\frac{{1}}{{2}}\right) = \frac{\pi}{{6}} \]