Proof for $\sin(x) > x - \frac{x^3}{3!}$

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $f(x)<f(\pi)<f(3)=3-4.5<-1$.

Solution 4:

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) $

Solution 5:

(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.