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