Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} =  0 \\ $
It seems that x = 0 is a critical point. 
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' =  -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?
 A: Take a decreasing sequence of positive real numbers $a_n$ such that $a_n\to 0$.
Now, consider the sequence $b_k=\sum_{n=1}^k (-1)^{n-1}a_n$. The alternating series criterion guarantee us that it converges to some $b$.
Note that $b_1=a_1$, $b_2=b_1-a_2\in(0,b_1)$, $b_3=b_2+a_3\in(b_2,b_1)$, etc. So the limit $b$ is lesser that the terms $b_{2k+1}$ and greater than $b_{2k}$.
Then, if $x<\sqrt 6$,
$$\sin x=\sum_{n=1}^\infty(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}>\sum_{n=1}^2(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}=x-\frac{x^3}{3!}$$
If $x\geq\sqrt 6$, the function $f(x)=x-x^3/6$ is decreasing and $f(\sqrt 6)=0$, so $f(x)<0$ for $x>\sqrt 6$. Since $\sin x>0$ for $0<x<\pi$, we have that $\sin x>f(x)$ for $0<x<\pi$. (Note that $\sqrt 6<\pi$).
Last, for $x\geq \pi$, $\sin x\geq -1$ and $f(x)<f(\pi)<f(3)=3-4.5<-1$.
A: You just have to prove your inequality when $x\in(0,\pi)$, since otherwise the RHS is below $-1$. Consider that for any $x\in(0,\pi/2)$,
$$ \sin^2 x < x^2 \tag{1}$$
by the concavity of the sine function. By setting $x=y/2$, $(1)$ gives:
$$ \forall y\in(0,\pi),\qquad \frac{1-\cos y}{2}<\frac{y^2}{4}\tag{2}, $$
so:
$$ \cos y > 1-\frac{y^2}{2} \tag{3} $$
for any $y\in(0,\pi)$. By integrating $(3)$ with respect to $y$ over $(0,x)$ we get our inequality.
A: Observe that:
$\\ \\ \displaystyle \sin(3\gamma)=\sin(2\gamma)\cos(\gamma)+\sin(\gamma)\cos(2\gamma)=2\sin(\gamma)\cos^2(\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=2\sin(\gamma)(1-\sin^2\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=3\sin(\gamma)-4\sin^3(\gamma)\Rightarrow \sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)$
\begin{equation}
    \sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)
\end{equation}
Do it $\displaystyle \gamma=\frac{\phi}{3^k}$:
\begin{equation}
    \sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Multiplying by $\displaystyle 3^{k-1}$:
\begin{equation}
    3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Applying summation on both sides of equality, we will have:
$\\ \\ \displaystyle \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\sum_{k=1}^{n}\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right) =\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\Rightarrow \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\\ \\$
Take the limit:
\begin{equation*}
        \lim_{n\rightarrow \infty}\sum_{k=1}^{n}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right)=\frac{1}{4}\left(\phi-\sin(\phi)\right)
\end{equation*}
On the other hand, using the inequality $ \displaystyle \sin x \leq x $ and  using the infinite  geometric progression formula, it follows that:
$\\ \displaystyle \sin\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi}{3^{k}}\Rightarrow \sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{3k}} \Rightarrow 3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{2k+1}}\Rightarrow  \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \sum_{k=1}^{\infty} \frac{\phi^3}{3^{2k+1}}=\frac{\phi^3}{3}\sum_{k=1}^{\infty} \frac{1}{3^{2k}}=\frac{\phi^3}{3\times 8}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \frac{1}{4}\left(\phi-\sin(\phi)\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \phi-\frac{\phi^3}{6}\leq \sin(\phi) $
A: A standard approach is to let $f(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$, and to show that $f(x)\gt 0$ if $x\gt 0$. 
Note that $f(0)=0$. We will be finished if we can show that $f(x)$ is increasing in the interval $(0,\infty)$.
Note that $f'(x)=\cos x-1+\frac{x^2}{2!}$. We will be finished if we can show that $f'(x)\gt 0$ in the interval $(0,\infty)$.
Note that $f'(0)=0$. We will be finished if we can show that $f'(x)$ is increasing in $(0,\infty)$.
So we will be finished if we can prove that $f''(x)\gt 0$ in the interval $(0,\infty)$.
We have $f''(x)=-\sin x+x$. Since $f''(0)=0$, we will be finished if we can show that $f'''(x)\ge 0$ on $(0,\infty)$, with equality only at isolated points. This is true. 
Or else for the last step  we can use the geometrically evident fact that $\frac{\sin x}{x}\lt 1$ if $x\gt 0$. 
Remark: It is more attractive to integrate than to differentiate, but we used the above approach because differentiation comes before integration in most calculus courses. 
For the integration approach, let $x$ be positive. Since $\sin t\lt t$ on $(0,x)$, we have $\int_0^x (t-\sin t)\,dt\gt 0$. Integrate. We get $\cos x+\frac{x^2}{2}-1\gt 0$ (Mean Value Theorem for integrals), so $\cos t+\frac{t^2}{2}-1\gt 0$ if $t\gt 0$. 
Integrate from $0$ to $x$. We get $\sin x+\frac{x^3}{3!}-x\gt 0$, or equivalently $\sin x\gt x-\frac{x^3}{3!}$. Nicer, by a lot.
A: (This started as a comment, but it is too big so I made it into an answer). 
That method is quite general, indeed. It will work if you do computations with enough care.
Somebody is suggesting the use of a Taylor polynomial. I think that with an approach like that, the only thing that you can prove easily is weaker than the one you are seeking. For example, to quickly prove that there exists a $\delta >0$ such that 
$$\tag{1}\sin x \ge \frac{1}{2}\left(x-\frac{x^3}{3!}\right), \qquad \forall x\in [0, \delta),$$
you can note that 
$$\frac{\sin x}{x-\frac{x^3}{3!}} = 1 +O(x^4), $$
so there exists a constant $C>0$ such that in a neighborhood of $x=0$ one has
$$\frac{\sin x}{x-\frac{x^3}{3!}}\ge 1 - Cx^4$$
and the right hand side is bigger than $\frac{1}{2}$ for $x\le \delta=\left( \frac{1}{2C}\right)^{\frac{1}{4}}$. 
Note that this value of $\delta$ is completely devoid of significance, it is just a small number that can in principle be very small. Also, in formula (1) you have a multiplicative factor of $\frac{1}{2}$ in the right hand side. All of that make this result way weaker than the one you are trying to prove; however, this proof is very quick and sometimes you do not really need the full-powered inequality. 
A: Check that
$$\int_0^x (x-t)^2 \sin^2\frac{t}{2} d\,t= \sin x - (x - \frac{x^3}{6})$$
WA link
A: Your solution is right. There is no problem or fault in your solution.
If you know something about trigonometric functions convergent series
Sin(x) can be written as x-x^3/6+x^5/5!.... which is >x-x^3/6.
For a beginner the method of yours is first to strike and it will always lead you to your result but if we know trigonometric functions convergent series we can conclude also
A: A trick is to get rid of the transcendental function.
As the sine is monotonic in $(0,\pi/2)$,
$$x>\arcsin\left(x-\frac{x^3}3\right),$$ and, as the two members coincide for $x=0$, by differentiation
$$1>\frac{1-x^2}{\sqrt{1-\left(x-\dfrac{x^3}3\right)^2}}.$$
Then it suffices that for $x<1$,
$$1-\left(x-\dfrac{x^3}3\right)^2>(1-x^2)^2,$$
$$-\frac{x^6}9-\frac{x^4}3+x^2>0,$$ which is true.
A: Recursive integration We know that $$0\le\cos a\le 1\implies \sin t = \int_0^t\cos s ds < t$$ for $0\lt t\lt z\lt x$. Integrating over $\color{blue}{(0,z)}$ we get
$$1-\cos z=\int_0^z\sin tdt < \int_0^ztdt= \frac{z^2}{2}$$
that is for all $0<z<x$ we have,
$$\color{blue}{1-\frac{z^2}{2}< \cos z\le 1}$$
integrating again over $\color{blue}{(0,x)}$ we get
$$\color{red}{x-\frac{x^3}{6} = \int_0^x 1-\frac{z^2}{2} dz< \int_0^x\cos z dz=\sin x}$$
that is
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x}$$

continuing with this process you get,
$$\color{blue}{1-\frac{x^2}{2}< \cos x< 1-\frac{x^2}{2}+\frac{x^4}{24} }$$
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x-\frac{x^3}{6} +\frac{x^5}{5!}}$$

More generally for  $n\geq1$, by induction we get
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}<\cos x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}+\frac{x^{4n}}{(4n)!} }$$
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!} <\sin x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{4n+1}}{(4n+1)!}}$$
