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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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    @MarcvanLeeuwen Yes, it is transformed (Not mentioned). Sorry.2012-10-23

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

$b>0\;\;\Longrightarrow\;\;|a|

So

$|\sin x|<\frac{1}{2}\Longleftrightarrow -\frac{1}{2}<\sin x<\frac{1}{2}\Longleftrightarrow \left\{\begin{array} {}-\frac{\pi}{6}

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.2012-10-23