Imagine you were told that in triangle \(ABC\), the side opposite \(C\) has length \(\quantity{5}{cm}\), the side opposite \(A\) has length \(\quantity{3}{cm}\) and angle \(A=\tfrac{\pi}{6}\). Think about how you would calculate angle \(C\).
At some point in your calculation you might obtain \(\sin C=\tfrac{5}{6}\), but what exactly does this tell you about \(C\)?
You may have heard of inverse trigonometric functions, such as \(\arcsin x\) or \(\sin^{-1}x\) or the phrase “inverse sine”. In principle, inverse trigonometric functions should be functions that undo the effects of functions such as \(\sin x\), \(\cos x\) and \(\tan x\). But we need to be careful because these functions are are not one-to-one; in fact they are periodic, so for instance there are infinitely many values of \(x\) that give the same value of \(\sin x\).
You might wonder where the names such as ‘arcsin’ come from.
Recall that in a circle of radius \(r\), the arc length subtended by an angle \(\theta\) radians has length \(r\,\theta\). So on a unit circle the length of an arc and the size of the angle which subtends it are equal. This is probably where the words come from.
Please note that \(\arcsin\) and \(\sin^{-1}\) mean the same thing and both are widely used in books and online, but note that \(\sin^{-1}x\) is not the same as \(\tfrac{1}{\sin x}\)! Similarly, \(\arccos\) means the same as \(\cos^{-1}\), and \(\arctan\) means the same as \(\tan^{-1}\). We’ll use \(\arctan\), \(\arccos\) and \(\arcsin\) in this resource.
Although we started with a situation where we were thinking about an inverse for \(\sin x\), we’ll think briefly about \(\tan x\) first before moving back to \(\sin x\) and \(\cos x\). Some of the issues that arise for an inverse of \(\tan x\) also arise for \(\sin x\) and \(\cos x\), but we’ll see that we have to be even more careful for these functions.
Recall from Making inverse functions that if we want to produce an inverse for a many-to-one function, we need to restrict the domain so that the function is one-to-one on its restricted domain. We also need the range of the restricted function to be the same as the range of the original function.
So for \(\tan x\),
How could we restrict the domain to make the function one-to-one whilst keeping its range unchanged?
Is there more than one way to do this?
The function \(\tan x\) is periodic with period \(\pi\). Restricting the domain to any one period will not change the range of the function.
From the graph of \(y=\tan x\) we can see that the function is always increasing. If we choose a domain between two adjacent asymptotes, such as the interval\(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), the function will be one-to-one and its range will still include all real numbers.
We could choose any interval of the form \(\left(\tfrac{(2n-1)\pi}{2}, \tfrac{(2n+1)\pi}{2}\right)\) for integer values of \(n\), but it seems most natural to choose the interval that includes zero.
The following graphs shows how we could restrict the domain of \(\tan x\) and the resulting function \(\arctan x\).
So although the equation \(\tan x = \sqrt{3}\) has infinitely many solutions, we say \(\arctan \sqrt{3}\) has the single value \(\tfrac{\pi}{3}\), and this is called the principal value for the equation \(\tan x = \sqrt{3}\). It is the value of \(x\) in the interval \(\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\) such that \(\tan x=\sqrt{3}\). The interval \(\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\) is known as the principal value range of the function \(\arctan\).
We can use the symmetry and periodicity of the \(\tan x\) graph to express all solutions of \(\tan x=\sqrt{3}\) in terms of \(\arctan \sqrt{3}\).
We’ll now return to thinking about \(\sin x\) and \(\cos x\).
The period of \(\sin x\) is \(2 \pi\), but within any interval of length \(2\pi\), there are two values of \(x\) that give the same value of \(\sin x.\) In other words, \(\sin x\) is not one-to-one or injective over its period. So in order to define an inverse we need to restrict to a smaller interval.
If we restrict the domain of \(\sin x\) to the interval \(\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\), then the resulting function is one-to-one and its range is the same as that of the unrestricted function. Its inverse is \(\arcsin x\) which has domain \([-1,1]\).
What is \(\arcsin 0.5\)?
What about \(\arcsin (-0.5)\)?
The situation for \(\cos x\) is very similar to that for \(\sin x\) but we need to restrict to a different interval.
If we restrict the domain of \(\cos x\) to \([0,\pi]\), then the range is still \([-1,1]\) and \(\cos x\) is now one-to-one. We use this domain to define \(\arccos x\) with domain \([-1,1]\) and range \([0,\pi]\). Note that both \(\cos x\) (restricted to this domain) and \(\arccos x\) are decreasing functions.
What is \(\arccos 0.5\)?
What about \(\arccos (-0.5)\)?
So \(\arctan a\), \(\arcsin a\) and \(\arccos a\) give the principal values for the equations \(\tan x=a\), \(\sin x=a\) and \(\cos x=a\) respectively, and we can work from these to find all possible solutions. You can see more about this in General solutions and in Inverse or not?.
Going back to our original triangle problem, we had \(\sin C=\tfrac{5}{6}\). The inverse function will give us the principal value (in radians) as \[C=\arcsin\tfrac{5}{6}\approx0.985.\] But there is more than one value of \(C\) that would satisfy our equation. We must always consider other possibilities when using inverse trigonometric functions.
Since we are talking about an angle in a triangle we need only consider values in the interval \([0,\pi]\). There is one more possible value which is \[\pi-\arcsin\tfrac{5}{6}\approx2.156.\]
