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

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?

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In general, when $w$ is a real number, $\sqrt{w^2}=|w|$ –  Thomas Andrews Oct 23 '12 at 13:14
"mentioned" := "transformed"? –  Marc van Leeuwen Oct 23 '12 at 13:22
@MarcvanLeeuwen Yes, it is transformed (Not mentioned). Sorry. –  lambda23 Oct 23 '12 at 13:24

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:

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

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I know how to change radians <-> degrees. Only the approving the inequality. Because I had this lesson and I don't know why it can be like that. Thank you for answer my confusion. –  lambda23 Oct 23 '12 at 13:29