# Why is the angle a well-defined quantity in a Euclidean space?

I am having a problem with an exercise. In a certain euclidean space, the angle between two vectors $u$ and $v$ is said to be $\theta$ which satisfies the equation:

$$\cos\theta=\frac{(u,v)}{\|u\|\|v\|}$$

Why is the angle a well-defined quantity if $\theta \in [0;\pi]$?

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It isn't, unless you also add the condition $0\le\theta\le\pi$. – Gerry Myerson Feb 10 '13 at 1:55
Yes we have this condition – Carpediem Feb 10 '13 at 1:55
@user43758: What does the graph of $\cos$ look like? – wj32 Feb 10 '13 at 1:58
If theta is between 0 and pi, cos is a decreasing function – Carpediem Feb 10 '13 at 2:00
Can someone help me? – Carpediem Feb 10 '13 at 2:03

It's not.

But suppose you add the restriction that $\|u\| >0, \|v\|>0$. By the Cauchy-Schwarz inequality, $|(u,v)| \leq \|u\|\|v\|$, so $$-1 \leq \frac{(u,v)}{\|u\|\|v\|} \leq 1.$$

It follows that $\theta\in [0,\pi]$ exists and is unique, since $\cos$ is a bijection from $[0,\pi]$ to $[-1,1]$.

$\hspace{3cm}$

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theta is between 0 and $\pi$ – Carpediem Feb 10 '13 at 2:05
Yes. Are you still confused about a part of the answer? – user7530 Feb 10 '13 at 2:07
Yes. In the question they already told be the interval between which theta is defined. Then they are asking me why is the angle a well defined quantity? – Carpediem Feb 10 '13 at 2:09
@user43758: It's well-defined because $\cos:[0,\pi]\to[-1,1]$ is a bijection (as stated in this answer). – wj32 Feb 10 '13 at 2:12
@user7530: I hope I didn't modify too much. – robjohn Feb 10 '13 at 11:45