Why $\sqrt{\sin^2 x}<0.5$ can be transformed in $|\sin x|<0.5$?

Question

Why $\sqrt{\sin^2 x}<0.5$ can be transformed in $|\sin x|<0.5$. Then $|\sin x|<0.5$ can be transformed in $-0.5<\sin x<0.5$? What is the proof of the inequality?

Answer 1

It's hard for me to guess what you mean by "be mentioned", but:

$$b>0\;\;\Longrightarrow\;\;|a|<b\Longleftrightarrow -b<a<b$$

So

$$|\sin x|<\frac{1}{2}\Longleftrightarrow -\frac{1}{2}<\sin x<\frac{1}{2}\Longleftrightarrow \left\{\begin{array} {}-\frac{\pi}{6}<x<\frac{\pi}{6}&\text{or}\\{}\\\;\;\;\frac{5\pi}{6}<x<\frac{7\pi}{6}\end{array}\right.$$

If you prefer degrees over radians remember:

$$\pi\,\text{rad.}=180^\circ\Longrightarrow \frac{\pi}{6}\text{rad.}=30^\circ\,\,\text{and etc.}$$