Proving |sinx| =< |x| for all real x.

Question

Everything I did to solve this problem matches the book, except for the last bit.

Use the following to verify the statement $|\sin x| \leq |x|$.

a) Show that for all $x \geq 0$, $f(x)=x-\sin x$ is non-decreasing.

$f'(x) = 1-\cos x$. Since $-1 \leq \cos x \leq 1$, we have $f'(x) \geq 0$ so that $f(x)$ is non-decreasing. Since $f(x)=0$ when $x=0$, we have $f(x)>0$ whenever $x>0$, and hence $\sin x \leq x$ when $x\geq 0$.

The book then says: And $|\sin x|=\sin x\leq x=|x|$. How can we say that $|\sin x|=\sin x$?

Answer 1

$|\sin x|=\sin x\leq x=|x|$ is of course wrong. So what is the correct argument?

For $\lvert x \rvert \ge 1$ we trivially have $|\sin x| \le|x|$.

Moreover we know that $\sin x\leq x = \lvert x \rvert$ for $x \ge 0$. It remains to consider $-1 \le x \le 0$. In this range $$\lvert \sin x \rvert = -\sin(x) =\sin(-x) \le -x = \lvert x \rvert.$$

Answer 2

Since both $|\sin x|$ and $|x|$ are even functions, you only need to show the statement when $x\ge0$. Also for $x>1$ the inequality holds true for obvious reasons and in the interval $[0,1]$ we have $$|\sin x|=\sin x\\|x|=x$$