# Prove $\sin(x) > x - \frac{x^3}{3!}$ on $(0, \sqrt{20})$

I'm having a bit of trouble with this because my attempted proof breaks down.

Proof: It is sufficient to show that $$f(x) = \sin(x) - x + \frac{x^3}{3!} > 0$$ on $$I = (0, \sqrt{20})$$. This is true if $$f'(x)$$ is strictly increasing on the interval and $$f(0) \geq 0$$. We can apply this property on the first and second derivatives as well. We note $$f^{(3)}(x) = -\cos(x) + 1$$. However I can only show that $$f^{(3)}(x) \geq 0$$ on $$I$$ since at $$x = \frac{3\pi}{2}$$ $$f^{(3)}(x) = 0$$

You are on the right track. The function $$f(x) = \sin(x) - x + \frac{x^3}{3!}$$ satisfies $$f(0) = f'(0) = f''(0) = 0$$ and $$f^{(3)}(x) = -\cos(x) + 1 \ge 0$$ with equality only at the points $$x_k = \frac \pi 2 + 2 k\pi$$, $$k \in \Bbb Z$$. It follows that $$f''$$ is strictly increasing on each interval $$[x_k, x_{k+1}]$$ Therefore $$f''$$ is strictly increasing on $$\Bbb R$$ and strictly positive on $$(0, \infty)$$.

Now you can conclude that $$f'$$ and consequently $$f$$ are strictly increasing on $$[0,\infty)$$, and therefore $$f(x) > 0$$ for all $$x > 0$$.

• Thank you very much! Apr 12, 2019 at 5:56

You can prove it directly: $$f(x)=\sin(x) - x + \frac{x^3}{3!}>0,x>0;\\ f'(x)=\cos (x) -1+\frac{x^2}{2}>0 \iff \left(\frac x2\right)^2>\sin ^2 \left(\frac x2\right) \iff \\ \left(\frac x2-\sin \frac x2\right)\left(\frac x2+\sin \frac x2\right)>0 \iff \\ \frac x2-\sin \frac x2>0 \iff \frac x2>\sin \frac x2,x>0$$ Note:

1) $$\cos (x)-1=-2\sin ^2 \left(\frac x2\right)$$.

2) $$x>\sin x, x>0$$ is true, because $$g(x)=x-\sin x$$ is strictly increasing.

Well, consider the series expansion of $$\sin(x)$$: $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$$ So from your definition of $$f(x)$$, $$f(x)=\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots\geq \frac{x^5}{5!}-\frac{x^7}{7!}$$ We can solve the inequality $$\frac{x^5}{5!}-\frac{x^7}{7!}>0\Leftrightarrow42-x^2>0\Leftrightarrow x<\sqrt{42}$$ since $$x>0$$.

So when $$0, your inequality is true.

I actually don't know where the $$\sqrt{20}$$ come from in the question. Anyone have a clue?

• Thank you. I believe the $\sqrt{42}$ comes from being less than $2\pi$. Apr 12, 2019 at 5:56

For $$x\ge0$$, integrating from $$0$$ at each step: $$\cos(x)\le1\\ \Downarrow\\ \sin(x)\le x\\ \Downarrow\\ 1-\cos(x)\le\tfrac12x^2\\ \Downarrow\\ x-\sin(x)\le\tfrac16x^3\\$$ Therefore, for all $$x\ge0$$, $$\sin(x)\ge x-\tfrac16x^3$$