How can I show that on $[-1,0]$ $|\sin(x)|\le|x|$? It's pretty obvious to me as I see the plot of the two functions but how can I prove it with some algebra?
 A: By mean-value theorem, if $-1 \leq x < 0$, then
$$
|\sin x - \sin 0| = |\sin x| \leq |x|\sup_{0 < t < x}|\cos t| \leq |x|.
$$
A: So here's an answer involving Calculus:
consider $f(x)=x-\sin x $
$f'(x) =1- \cos x > 0$
for $x \in [0,1]$, $f$ is increasing, so $f(x)>f(0)=0$ for $x \in [0,1]$
which means $x>\sin x \implies |x|>|\sin x|$ since $x$ and $\sin x$ is positive here.
for $x \in [-1,0]$, $f$ is increasing, so $f(x)<f(0)=0$ for $x \in [-1,0]$
which means $x<\sin x \implies |x|>|\sin x|$ since $x$ and $\sin x$ is negative here.
Therefore $|x|>|\sin x|$ for $x \in [-1,1]$
A: (image taken from robjohn's answer here)
A geometrical proof can be seen in the following image:
$\sin x$ is the height from $C$ which is smaller than $BC$, which, in turn is smaller than the length of the arc $BC$, which is equal to $x$.

A: Use the mean value theorem:
There is some number $c$ between $x$ and $0$ for which
$$
\sin' c = \frac{\sin x - \sin 0}{x - 0}.
$$
Since $|\sin'c|=|\cos c|\le 1$, we have
$$
\left|\frac{\sin x} x \right| \le 1,
$$
so
$$
|\sin x| \le |x|.
$$
There is no need to assume $x$ is in $[-1,0]$ to get this conclusion.
