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I have this exercise with an unclear answer. The question is this: $$\arccos x = -\frac{\sqrt3}{2}\,.$$

The answer is this: $$\begin{gather*} \varphi(x)= \arccos x\\ V_\varphi = [0,\pi]\\ -\frac{\sqrt3}{2}\notin V_\varphi\,. \end{gather*} $$

can someone this a bit more to me? How does this prove that there are no solutions?

David Richerby
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Johan Hjalmarsson
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  • Is the question "Does it exist a (real) $x$ such that arccos$(x)=-\frac{\sqrt{3}}{2}?$" ? – MattAllegro Oct 25 '14 at 11:49
  • No need to limit yourself to real $x$. There's no complex $x$ with that property either. (In both real and complex numbers, that is true if you take the standard primary branch.) – Christopher Creutzig Oct 25 '14 at 13:17

3 Answers3

8

As a couple of others have already pointed out, $-\sqrt{3}/2$ is simply not in the range of the arccosine. Here's an explanation as to why that's true.

Here's the graph of the cosine function over the interval $[-\pi,2\pi]$:

enter image description here

The issue is that this function is not one-to-one. As a result, we must restrict it to an appropriate domain where it is one-to-one in order to talk a about a restricted inverse function. Conventionally, the interval $[0,\pi]$ is chosen, which yields something like the following:

enter image description here

Now, the inverse of this restricted version of the cosine is what we know as the arccosine and its graph looks like so:

enter image description here

Of course, the domain and range have flipped - thus, $-\sqrt{3}/2$ (or any other negative number) is not in the range!

Mark McClure
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6

Domain of $\arccos(x)$ is $[-1,1]$ (not $\mathbb{R})$, so the range of values is $[0,\pi]$, but $\frac{-\sqrt{3}}{2}<0$, so it isn't in range.

agha
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4

Hints:

Do you know that the range of $f(x)=\arccos x$? It is not $\Bbb R$.

Paul
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