Trigonometric Inequalities-Proof Verification. Prove the following inequalities:
If $x\in\Bbb R$, $x\ge0$ then,
(1) $-x\le \sin x\le x$. Moreover, does this inequality hold if $x<0$
(2) $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1$
(3) $x-\displaystyle\frac{x^3}{6}\le \sin x\le x$
(4) $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$
(5) Also show that, $\left|\sin x\right|\le | x|$, $\forall x\in \Bbb R$
My Attempt:
(1)
Since, $\displaystyle sin^2t+cos^2t=1$,  $ \forall \;\;t\in \displaystyle\Bbb R $
$\;\;\;\;\;\;\;\;\implies$ $\;-1\le \cos t\le1$ $ \forall \;\;t\in \displaystyle\Bbb R $
If $x\ge0$ then, 
$$\int_{0}^x-dt\le\int_{0}^x \cos t\;dt\le\int_{0}^x dt\;\;\implies  -x\le\sin x\le x$$ 
If $x<0$ we have,  $x\le \sin x\le -x$. 
(2)
On integrating the inequality in (1), we get 
$$\int_{0}^x-t\;dt\le\int_{0}^x \sin t\;dt\le\int_{0}^x t\;dt\;\;\implies -\frac{x^2}{2} \le - \cos x +1 \le   \frac{x^2}{2} \implies 1-\frac{x^2}{2} \le \cos x$$
Therefore we can conclude,  $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1$
On proceeding similarly, I have verified that (3) and (4) can also be obtained.
Is the above approach correct?
Furthermore I am curious to see what happens in (2), (3) and (4) for $x\le 0$.
In (2), 
If $x \le 0$  $\implies $ $-x \ge 0$
Therefore,  $1-\displaystyle\frac{(-x)^2}{2}\le \cos (-x)\le 1$ $\implies $  $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1$
Thus,  $$1-\displaystyle\frac{x^2}{2}\le \cos x\le 1, \;\; \forall x\in \Bbb R$$
In (3), 
If $x \le 0$  $\implies $ $-x \ge 0$
Therefore, $(-x)-\displaystyle\frac{(-x)^3}{6}\le \sin(-x)\le (-x) \implies -x+\displaystyle\frac{x^3}{6}\le -\sin x\le -x \implies x \le \sin x\le x-\displaystyle\frac{x^3}{6}$
Thus, $$x-\displaystyle\frac{x^3}{6}\le \sin x\le x \;\;, \forall x\ge 0 $$
      $$x \le \sin x\le x-\displaystyle\frac{x^3}{6}\;\;, \forall x\le 0$$
Similarly for part (4) ,  $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}, \;\; \forall x\in \Bbb R$
How do I go about with proving (5)?
Have I done something wrong?

Everything looks fine. For (5) use (1) and the fact that $|a| \le |b|$ equivalent $-|b| \le a \le |b|$. 
For $x \ge 0$ then, using (1) $|sin(x)|\le x$, therefore $|sin(x)|\le |x|$. 
For $x \lt 0$ then, using (1) $x \le sin(-x) \le -x$ equivalent $x \le -sin(x) \le -x$ equivalent $x \le sin(x) \le -x$ equivalent $-|x| \le sin(x) \le |x|$ 
