Proof about boundedness of $\rm Si$

Question

$\def\Si{{\rm Si}}$ I want to prove the boundedness of $$\Si(x) := \int_0^x \frac {\sin \xi} \xi d\xi$$ as part of a homework (about the non-surjectivity of $\mathcal F : L^1(\mathbb R) \to C_0^0(\mathbb R)$). As a first step, I found (by means of plotting) $$\frac{\sin x}{x} \leq \cos \frac x 2 \qquad \forall\ x \in (0, \pi)$$ but couldn't easily prove that. If that is proved, we had, including some other arguments $$\Si(x) \leq \int_0^\pi \frac {\sin \xi} \xi d\xi \leq \int_0^\pi \cos\frac\xi 2 d\xi = 2$$ And I'd be done.

How can I prove $$\frac{\sin x}{x} \leq \cos \frac x 2 \qquad \forall\ x \in (0, \pi)$$ in a simple way, or do you have any alternative proof for my objective?
Thanks in advance

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Okay with the hints provided by @HaraldHanche-Olsen in his answer, it became very clear:

$$\frac{\sin x} x \stackrel{\text{addition thm.}}{=} \frac{2 \sin\frac x 2 \cos\frac x 2}{x} = \underbrace{\frac{\sin\frac x 2}{\frac x 2}}_{\leq 1} \cos \frac x 2 \leq \cos\frac x 2$$

Answer 1

Here are two hints for the price of one: $$\frac{\sin x}{x}\le 1$$ (I am sure you know that one) and $$\sin x=2\sin\frac x2\cdot\cos\frac x2.$$

Answer 2

Here is an alternate proof. There is no trouble from $0$ to $1$, so let's integrate from $1$ on.

To find $\int_1^x \frac{\sin t}{t}\,dt$, use integration by parts. Let $u=\frac{1}{t}$ and $dv=\sin t\,dt$. Then $du=-\frac{1}{t^2}\,dt$ and we can take $v$ to be $-\cos t$. It follows that $$\int_1^x \frac{\sin t}{t}\,dt=\left.-\frac{\cos t}{t}\right|_1^x +\int_1^x \frac{\cos t}{t^2}\,dt.$$

Everything is fine now. The integral of $\frac{\cos t}{t^2}$ is no problem, the absolute value of the function is bounded above by $\frac{1}{t^2}$.

Remark: The above idea, perhaps iterated a few times, is a useful device for making estimates.