Prove $\max \cos(x)$ is $1$ and $\min \cos(x)$ is $ -1$ Prove $\max \cos(x)$ is $1$ and $\min \cos(x)$ is $-1$
How to prove it with only calculus and not multivariable calculus?
Please notice that this is not a homework question, but a pre-exam question. Thanks a lot.
 A: Since
$$\cos^2 x+\sin^2 x=1$$
then
$$-1\le\cos x\le 1$$
and we have
$$\cos(0)=1\quad;\quad \cos(\pi)=-1$$
Conclude.
A: Try the geometric definition of cosine:  $|\cos \theta|=\frac{|\text{adj}|}{|\text{hyp}|}$.  Since the hypotenuse is the longest side in a right triangle, we find that the ratio is at a maximum when $|$adj$|=|$hyp$|$.  We can also find an instance of this happening, that is when $\theta=0$, or there is no opposite side.  Therefore $$|\cos \theta|\le 1\Longrightarrow -1\le \cos\theta\le 1$$
A: Put $f(x) = \cos x - 1 $. Then $f' = - \sin x $.
$$ f' = 0 \iff \sin x = 0 \iff x = n \pi $$
and $f'' = - \cos x \implies  f''(\pi n) = 1 $ if $n$ is odd and $= -1$ if $n$ is even.
So, if $n$ is even, then $x = \pi n$ is a max and 
$$ f(\pi n) \geq f(x) \implies 0 \geq \cos x - 1 \implies 1 \geq \cos x $$ 
Similarly, if $n$ is odd, then $x = \pi n $ is a min and hence 
$$ f(x) \geq f(\pi n) \implies \cos x -1 \geq -1 -1 \implies \cos x \geq -1$$
$$ \therefore 1 \geq \cos x \geq -1 $$
A: It is $\cos{(\varphi)}=\Re{({e^{i\varphi}})}$ and $|e^{i\varphi}|=1$.
$|\Re{(z)}|\leq|z|$ together with the evaluation at $0$ and $\pi$ proves your question.
