# Inverse Trig and infinite values (arccos)

I understand that trig ratios can have infinite values for the same value of $x$

$\cos(x)$ for example. Since $\cos(x)$ shows the relationship between two sides of a triangle and that ratio can have an infinite amount of combinations.

IE $\cos(x)$ where ${ x = \pi}$, we get $-1$, or when ${ x = 3\pi}$, we get $-1$. But what about the inverse cos function?

If we have ${ \cos^{-1}(0.5)}$ this would be ${ \pi/3}$, but I'm having a disagreement with someone about whether this can also have an infinite amount of values. I think it can't, they think otherwise. I mentioned that the function ${ \cos^{-1}(x)}$ has a domain where ${ -1 \le x \le 1}$, however they said that if we were looking at functions i'd be correct however looking for only the value we can have other (infinite) answers where

\begin{align} \cos^{-1} (0.5) &= \frac{\pi(6n-1)}{3} \end{align}

Where $n$ is a real integer.

But I disagree. Entering this in any scientific calculator returns error.

Could someone point me in the right direction?

You're correct, in that cos(x) is not a one-to-one function; there are infinite values of x for every value of y. This will reflect on the inverse cos function as well.

Indeed, sometimes this needs to be utilized to get the correct answer to a problem.

However, the arccos you'll find in a calculator is by definition restricted to the range of 0 < x < $\pi$, so it behaves like a function. This is necessary to do so we can perform calculus on it.

So, you're right in that inverse cos of any value has infinite "inputs", but your friend is right in that inverse cos will only ever return one of those inputs. Which of these two is actually important to the problem depends on that problem.

You can consider $\arccos: [-1,1]\to [0,\pi]$ or $\arccos: [-1,1]\to [r,r+\pi]$ for any $r>0.$ Once you fixed the target there is only one value for $\arccos x.$ But if you don't fix the target you have infinite values. In some situations (for example, if you are working with angles in the second quadrant) you must need an answer that belongs to $[\pi/2, 3\pi/2].$ That is, it depend on the context you are dealing with.

If you want $\cos^{-1}(x)$ to be a function, then it should only have one result for any given element in the range $[-1, 1]$. This is because in order for a rule to be considered a function, it must have only one unique output for any given input.

It is true that there are an infinite number of angles whose cosine is equal to $\frac{1}{2}$. However, we usually define the inverse cosine function to map to an interval such as $[0,\pi]$, so that we know the result must live in this interval. In this way, we ensure that there is a unique result for every input value in $[-1,1]$.

Now, our choice of taking the range to be $[0,\pi]$ is not forced on us. Maybe in a particular situation you are interested in which angle in the range $[-\pi / 6, 5\pi/6 ]$ has a cosine of $\frac{1}{2}$. In this case, the angle you are looking for is given by the inverse cosine function whose range is that interval.

So, yes there are an infinite number of angles whose cosine is $\frac{1}{2}$. However,to ensure that the inverse cosine is indeed a function, you should restrict it's range so that there is only one unique angle.