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By using Taylor formula, prove that for every $x$ in the closed interval $[0, \pi/2]$ we have:

$$ \sin x \le x - \frac{x^3}6 + \frac{x^5}{120} $$

I know that these are first terms of the sin expansion, but can't understand why it is enough to prove the inequality.

(Excuse me for the lack of knowledge in writing the text in the proper way)

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    Please take a look at [MathJax guide](http://math.stackexchange.com/help/notation) to learn how to format maths here. Your question is unclear to me, are you asking why is it enough to prove inequality to solve the problem? It is trivial, isn't it? So what exactly do you want to know? – 2017-01-02
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    It does not appear that trivial to me unfortunately. I only used taylor in limits. I mean, Taylor is an approximation right? So how do I know that in this case the approximation is precise enough that the inequality holds? – 2017-01-02
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    The problem states something like: "Use Taylor expansion to prove inequality" and you basically asked "Why is it enough to prove inequality to prove inequality?" which I'm guessing is not what you wanted to ask. – 2017-01-02
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    What is the relation between taylor and an inequality? How can I say that the taylor approximation is good enough? Anyway, I feel I am not that good at explaining myself sadly. Could you gladly just write an answer to the problem in a rigorous way ignoring my comments thereafter? – 2017-01-02

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Note that there exists a number $\xi \in (0,\pi/2)$ such that

$$\sin(x)=x-\frac16x^3+\frac1{120}x^5-\frac{1}{7!}\cos(\xi)x^7 \tag 1$$

Since $\cos(\xi)>0$ for $\xi \in (0\,\pi/2)$, the last term in $(1)$ is negative and hence

$$\sin(x)\le x-\frac16x^3+\frac{1}{120}x^5$$

for $x\in [0\,\pi/2]$.

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    @user403565 You're welcome. My pleasure. And Happy New Year! -Mark – 2017-01-02
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Let we set $I=\left[0,\frac{\pi}{2}\right]$. We have: $$ \forall x\in I,\qquad \sin(x)\leq x \tag{1} $$ hence by integrating both sides over the interval $[0,z]$, $$ \forall z\in I,\qquad 1-\cos(z) \leq \frac{z^2}{2}\tag{2} $$ hence by integrating both sides over the interval $[0,x]$, $$ \forall x\in I,\qquad x-\sin(x) \leq \frac{x^3}{6} \tag{3} $$hence by integrating both sides over the interval $[0,z]$, $$ \forall z\in I,\qquad -1+\frac{z^2}{2}+\cos(z) \leq \frac{z^4}{24}\tag{4} $$ hence by integrating both sides over the interval $[0,x]$, $$ \forall x\in I,\qquad -x+\frac{x^3}{6}+\sin(x) \leq \frac{x^5}{120} \tag{5} $$ as wanted.