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?
Why \sqrt{\sin^2 x}<0.5 can be transformed in |\sin x|<0.5?
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$\begingroup$
trigonometry
inequality
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0@MarcvanLeeuwen Yes, it is transformed (Not mentioned). Sorry. – 2012-10-23
1 Answers
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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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0I 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