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I need to understand this because I think that I don't know the meaning of arc...

$t=\cos(x) \Rightarrow x=\arccos(t)$ ??


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Not quite true. Let $x=2\pi$. Then $\cos x=1$, but $\arccos(1)=0$. Many similar exmples. But it is true if $0\le x\le \pi$. – André Nicolas Mar 1 '14 at 22:40
Others have given nice descriptions of how to make this claim completely precise. But the basic point is that this is the definition of $\arccos$: it's the function that gives you an angle which you can take cosine of to get back to your original number. Such a function doesn't exist in the simple way I've described it, so you have the details in the other comments and answers. – Kevin Carlson Mar 1 '14 at 23:12
By definition, $\arccos(\cos(x)) = x$ – okarin Mar 2 '14 at 1:45

Actually, the truth is slightly more complicated than that. The problem is that if I tell you that $\cos x = t$, I have not yet given you enough informations for you to reconstruct what $x$ is. For example, $x=\frac{\pi}{2}$ and $x=-\frac{\pi}{2}$ both satisfy the equation $\cos x = 0$. In general, therefore, it is not possible to define an inverse of the cosine function.

All is not lost, however, as the cosine function is bijective on the domain $[0,\pi]$. On this interval, it has an inverse, and it is this inverse that we call the $\arccos$ function. This means that, for $x\in [0,\pi]$, you define $\arccos t$ as "the value $x$ for which $\cos x = t$".

This allows you to solve any general equation $\cos x = t$. It immediatelly gives you one solution, $x=\arccos t$. Because $\cos$ is an even function, you immediatelly know that $x=-\arccos t$ is also a solution, as $\cos(-\arccos t)=\cos(\arccos t)=t$. Now, knowing that $\cos$ is a periodic function with period $2\pi$, you also know that if $x$ is a solution, then the value $x+2k\pi$ is a solution for any integer $k$. This means that all solutions are the following:

$$x=\arccos t + 2k\pi, k\in\mathbb Z\\ x=-\arccos t + 2k\pi, k\in\mathbb Z.$$

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Did I make a mistake in my answer? Why did I deserve a downvote? – 5xum Mar 1 '14 at 23:17
v good answer, I'd not really thought about this before properly. (You might want to say what bijective means?) – TooTone Mar 2 '14 at 0:18
A function is bijective if it is surjective and injective. It is surjective if it hits all values (like $\cos$ which hits all values between $-1$ and $1$) and injective if it doesn't hit any values twice (the $\cos$ function on the whole real numbers is not injective, because $\cos(0)=\cos(2\pi)$. Another example of a function which is not injective is $x^2$). This means a bijective function is a function that hits each value exactly once. The important thing is that bijective functions are exactly the functions which have an inverse. – 5xum Mar 2 '14 at 0:22

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