Prove $\cos(x)$ always smaller than $1$ for all values of $x$!!! In fact this question was ask in a linear algebra section!
The original question is "why is $|\cos(x)|$ never grater than one?" if $x$ is $90$ then I can prove it using algebra as follows!
if the length of two side of a right angled triangle are $x$ and $y$
then $\cos(x)=\frac{x}{\sqrt{x^2+y^2}}$ and this is always smaller than one
But how can i prove the above statement, if the angle is smaller or grater than 90 by considering this vector equation
$\left \|v-w  \right \|^{2}=\left \| v \right \|^2+\left \| w \right \|^2  -2\left \| v \right \|\left \| w \right \|\cos(x)$ 
where $v$ and $w$ are two vectors! 
Thanks
 A: Hint:
Well, since:
$$\lVert v\rVert^2+\lVert u\rVert^2-2\lVert v\rVert\lVert u\rVert\cos x= \lVert u-v\rVert^2\geq0$$
We also have that:
$$\lVert v\rVert^2+\lVert u\rVert^2-2\lVert v\rVert\lVert u\rVert\cos x\geq0\tag{$\star$}$$
So, we must have that:
$$\Delta=(-2\cos x)^2-4\leq0$$
Viewing $(\star)$ as a quadratic with respect to $\lVert v\rVert$.
Complete solution: Now it is evident that:
$$4\cos x^2-4\leq0\Leftrightarrow\cos^2x\leq1\Leftrightarrow|\cos x|\leq1$$
A: It depends on your definition of $\cos$.


*

*If you use geometry to define $\cos \theta$ for $0 \le \theta \le \frac{\pi}{2}$ and then extend it to all real $\theta$. Then it's an easy geometry problem.

*If you define $\cos$ in linear algebra by
$$\cos \theta=\frac{u\cdot v}{|u||v|}$$
then it can be proved by Cauchy-Schwarz, which is basically Βασίλης Μάρκος's proof.

*If you define it in analysis using
$$\cos \theta=\sum_{n=0}^\infty \frac{\theta^{2n}}{(2n)!}$$
then you can at the same time define
$$\sin \theta=\sum_{n=0}^\infty \frac{\theta^{2n+1}}{(2n+1)!}$$
You then have to show that the two series converges for all $\theta$.
After that, it is easily to show that
$$\frac{d}{d\theta}\sin \theta = \cos \theta$$
$$\frac{d}{d\theta}\cos \theta = - \sin \theta$$
Then we have
$$\frac{d}{d\theta}\left(\sin^2 \theta + \cos^2 \theta\right)=2\sin \theta \cos \theta - 2\cos \theta \sin \theta =0$$
Hence
$$\sin^2\theta + \cos^2 \theta= \sin^2 0 + \cos^2 0 = 0 + 1 = 1$$
for all real $\theta$.
Therefore
$$|\cos \theta| \le 1$$
for all real $\theta$.
By the way, I find it interesting to prove that definition (1) and (3) are consistent. In other words, to show that these newly defined functions we learn in University are indeed the old functions we learned in high schools. So let me prove it here as an aside.
From (3), let us define $\arccos$ and $\arcsin$ as the inverse functions of $\cos$ and $\sin$, respectively.
Then if we restrict ourselves to angle between $0$ and $\pi/2$ and let
$$y=\arcsin x$$
we have
$$\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^2}}$$
Now the equation of a unit square is
$$x^2+y^2=1$$
and hence the arclength is
$$\theta=\int_0^y\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy=\int_0^y\frac{dy}{\sqrt{1-y^2}}=\arcsin y - \arcsin 0=\arcsin y$$
Hence
$$y=\sin\theta$$
consistent with the geometrical definition in (1).
A: We all know about angles in the plane, and every high school student learns that 
$\tag 1 -1 \le cos(\theta) \le +1$
for any angle.
OK, so now we are told that using the dot product we can define the angle $\theta$ between two vectors $\vec u$ and $\vec v$ emanating from the zero vector in a finite dimensional space like $\mathbb R^n$. Wow, that is certainly neat!
Here is the formula for $\theta$,

So, to get the angle we have to know about $arccos$, the inverse of the the $cos$ function. The beauty is that it all works. The question that the OP is attempting to answer does not take one on an 'enlightened path'. It is more fruitful and interesting to ask why

can only take values in the closed interval $[-1,+1]$. And as mentioned in a comment by velut luna, you can draw inspiration from the Cauchy-Schwarz inequality.
A: Let's take the vector $\vec v$, now let's rotate it by $\theta$, we will call this vector $\vec v_\theta$. Now let's call the rejection from $\vec v_\theta$ to $\vec v$ to be $\mathbf v_{⊥v}$ now we have $|\cos\theta| =\frac{\|\mathbf v_{⊥v}\|}{\|\vec v\|}$, the rejection is part of $\vec v$, which means it is shorter or equal to $\vec v$, so we get that $|\cos\theta| =\frac{\|\mathbf v_{⊥v}\|}{\|\vec v\|}\le 1$. Of course this is not the most formal way, but I find it very intuitive
