# Calculus: $\arccos(x)$ - task problem

I´m having a hard time figure how to calculate the exact result on a task like this $$\arccos(\cos(-7\pi/6))$$ Where do I start here? Any tip or help would be greatly appreciated.

-

The principal branch of $\arccos(x)$ usually returns a number between $0$ and $\pi$. So when you evaluate $\arccos(\cos(-7\pi/6))$, you will get the value $y$ between $0$ and $\pi$ for which $\cos(y)=cos(-7\pi/6)$. Since $-7\pi/6 \equiv 5\pi/6 \mod 2\pi$, the answer is $5\pi/6$. (The relation "mod" is an equivalence relation used for example when some properties are repeated. Most common cases are the remainder when doing division. So we'd say $5\equiv3\equiv 1\mod 2$ because $5=2\times 2+1,3=2+1$)

To see this, imagine a clock with $12$ hours (hard right!), but instead of being $1,2,\ldots 12$ o'clock, your clock shows $\pi/6,2\pi/6,\ldots 12\pi/6$.

Your clock also has the weird functionnality of going counterclockwise, and it's $12$ o'clock is at the usual $3$ o'clock. So if I say it's $5\pi/6$ o'clock, then imagine me going $5$ hours on this weird clock. If I say its $13\pi/6$ o'clock, well I've done a whole turn plus one hour, so it is equivalent to $1\pi/6$ o'clock.

When we say $-7\pi/6$, it is like if you want to know what time it was $7$ hours ago, so on this weird clock, you go clockwise $7$ hours. You can convince yourself with a drawing such as.

-
Thank you! Is there chance you could evaluate "Since −7π/6≡5π/6 mod2π"? What does mod stand for? – marsrover Nov 4 '12 at 22:34
I'm working on something for you – Jean-Sébastien Nov 4 '12 at 22:56
Cosine is periodic, with period $2\pi$. Since $-7\pi/6$ and $5\pi/6$ differ by $2\pi$, they have the same cosine. – Gerry Myerson Nov 4 '12 at 22:59
Wooow, this is truly amazing Jean, many thanks! – marsrover Nov 5 '12 at 20:24