Show that $\sin x$ lies between $x-x^3/6$ and $x \;$ $\forall x \in R$ 
Show that $\sin x$ lies between $x-x^3/6$ and $x \;$ $\forall x \in R$

I am getting:
$$\sin(x) = x - \frac{x^3}{3!} + R_4(x)$$
where $R_4(x) = \frac{\cos(c)x^5}{5!}$ for some $c$ between $0$ and $x$ 
I want to prove $R_4(x)\geq 0$ to arrive at the result $x-x^3/3 \leq \sin(x)$.
for:$$0 \leq x \leq \pi/2 \Rightarrow 0<c<\pi/2 \Rightarrow R_4(x)\geq 0$$ but for: $$-\pi/2 \leq x < 0 \Rightarrow -\pi/2 < c < 0  \Rightarrow R_4(x) < 0$$
How can I proceed with this ? There are many cases that I need to check

The aim is to show that:
$$ x - \frac{x^3}{3!} \leq \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + ...\leq x $$
For $sin x\leq x$ it suffices to use MVT:
$$\cos c =\frac{\sin x- \sin 0}{x-0}=\frac{\sin x}{x}\implies -1\leq \frac{\sin x}{x}\leq \implies \frac{\sin x}{x}\leq 1 \implies sin x\leq x$$
For  $x - \frac{x^3}{3!}\leq \sin(x) $ see here
Proof for $\sin(x) > x - \frac{x^3}{3!}$
