# I am confused why cos (arccos 3π) has an answer of undefined. I see that arccos (cos 2π) is outside the range too, but you end up with arccos(cos 2π)=0? Why wouldn't the second one be undefined too?

Aug 23, 2016

The cosine function has a range of $\left[- 1 , 1\right]$, meaning the inverse cosine function has a domain of $\left[- 1 , 1\right]$. The reason $\cos \left(\arccos \left(3 \pi\right)\right)$ is undefined is because $\arccos \left(3 \pi\right)$ is undefined, as $3 \pi$ is outside of the domain for $\arccos$.

To put it another way, $\arccos \left(x\right)$ is a value such that $\cos \left(\arccos \left(x\right)\right) = x$. If $| x | > 1$, then no such value exists within the real numbers.

Note in the following graph $y = \arccos \left(x\right)$ that $\arccos$ is only defined for $- 1 \le x \le 1$.

graph{arccos(x) [-2, 2, -1, 4]}

There is no problem for the second one, as $\cos \left(2 \pi\right) = 1$, which is within the domain of $\arccos$. $\arccos \left(\cos \left(2 \pi\right)\right) = \arccos \left(1\right)$, which is a value such that $\cos \left(\arccos \left(1\right)\right) = 1$.

Because there are infinitely many values $x$ such that $\cos \left(x\right) = 1$, we need to make some restriction to make $\arccos \left(1\right)$ produce a unique result. The restriction used is $0 \le \arccos \left(x\right) \le \pi$. Note again within the above graph that the maximum value for $y$ is $\pi$ and the minimum value for $y$ is $0$.

Within the interval $\left[0 , \pi\right]$, the only value $x$ such that $\cos \left(x\right) = 1$ is $x = 0$. Thus, we see why the second result occurs:

$\arccos \left(\cos \left(2 \pi\right)\right) = \arccos \left(1\right) = 0$