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?
Best Answer
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.