Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

1 Answer 1

up vote 3 down vote accepted

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:

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.