Prove that $|\sin x-x |\le \frac{1}{6}|x|^3$

Question

Prove that $|\sin x-x |\le \frac{1}{6}|x|^3$

I've been trying to solve this problem for the last hour but I just can't crack it. I know I'm supposed to use Cauchy's Mean Value Theorem somehow but can't figure out how.

Edit: This problem is from chapter 4 of Foundations of Analysis by Taylor. The book hasn't covered Taylor series yet. I am supposed solve this using a method that is already covered. It is very likely that the intended solution involves the usage of Cauchy's form of the mean value theorem.

Answer 1

A solution using Cauchy's Mean-Value Theorem: With $$ f(x) = \sin(x) - x, g(x) = x^3 $$ we have for all $x > 0$ $$ \frac{\sin(x)-x}{x^3} = \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{f'(c)}{g'(c)} = \frac{\cos(c)-1}{3c^2} $$ for some $c$ between $0$ and $x$. Now apply Cauchy's Mean-Value Theorem again twice to conclude that this is $$ \begin{align} &= \frac{-\sin(d)}{6d} \quad \text{for some $d$ between $0$ and $c$} \\ &= \frac{-\cos(e)}{6} \quad \text{for some $e$ between $0$ and $d$} \end{align} $$ and therefore $$ \left| \frac{\sin(x)-x}{x^3} \right| = \frac{|\cos(e)|}{6} \le \frac 16 \, . $$

Answer 2

Assume $x \ge 0$.

Set $f(x) = \sin x-x$. We have $f'(x) = \cos x-1 \le 0$ so $f$ is decreasing on $[0,+\infty)$. Therefore $f(x) \le f(0) = 0$ so $\sin x \le x$.

Set $$g(x) = 1-\cos x - \frac12x^2$$ We have $g'(x) = \sin x -x\le 0$ so $g$ is decreasing on $[0,+\infty)$. Therefore $g(x) \le g(0) = 0$ so $1-\cos x \le \frac12x^2$.

Finally, set $$h(x) = x-\sin x - \frac16x^3$$ We have $$h'(x)= 1-\cos x - \frac12x^2 \le 0$$ so $h$ is decreasing on $[0,+\infty)$. Therefore $h(x) \le h(0) = 0$ so $x-\sin x \le \frac16x^3$.

Answer 3

When $0\leq x\leq4$ then $$\sin x-x=-{x^3\over6}+{x^5\over120}-{x^7\over5040}+\ldots\ ,$$ and therefore $|\sin x-x|\leq{x^3\over6}$ since the series on the RHS is alternating, and in the given $x$-domain the terms are decreasing in absolute value. When $x\geq4$ we can estimate $$|\sin x-x|\leq 1+x\leq{1\over6}x^3\ ,$$ since the rightmost term is obviously abounding when $x\geq4$.