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Thread: inverse sin etc.

  1. #1
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    inverse sin etc.

    Ok i solved these problems.

    a) $\displaystyle sin(arcsin(-\frac{1}{2}))$

    $\displaystyle -\frac{1}{2}$ belongs to interval $\displaystyle [-\frac{\pi}{2]}, \frac{\pi}{2]}]$

    so solution is $\displaystyle -\frac{1}{2}$

    b) $\displaystyle arcsin(sin(\frac{7\pi}{6})$

    $\displaystyle \frac{7\pi}{6}$ belongs to interval $\displaystyle \Re$
    so solution is $\displaystyle \frac{7\pi}{6}$

    c) $\displaystyle tg(arctg(-1))$

    -1 belongs to $\displaystyle /Re $

    so solution is -1
    Last edited by Fabio010; Nov 12th 2011 at 09:32 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Confirm if solutions are correct please.

    a) and c) are correct but you can just use the fact that $\displaystyle \sin(\arcsin(x))=x$ and $\displaystyle \tan[\arctan(x)]=x$ (if you know why)
    For b) first of all $\displaystyle \sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=\frac{-1}{2}$
    So that means $\displaystyle \arcsin\left(\frac{-1}{2}\right) \in \[\frac{-\pi}{2},\frac{\pi}{2}\] $

    So that means $\displaystyle \frac{7\pi}{6}$ can't never be an answer because it's not in $\displaystyle \[\frac{-\pi}{2},\frac{\pi}{2}\] $
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  3. #3
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    Re: Confirm if solutions are correct please.

    damn im confusing the domains

    so for sin(arcsin) = x and arcsin(sin x) = x

    x need to be in [-pi/2 , pi/2 ]
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Confirm if solutions are correct please.

    You can't say $\displaystyle \arcsin[\sin(x)]=x$, question b is an example of why it's not true (also think about it why it's not true in general).
    Actually I have given the answer to question b) with this quote:
    Quote Originally Posted by Siron View Post
    $\displaystyle \sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=\frac{-1}{2}$
    That means $\displaystyle \arcsin\left(\frac{-1}{2}\right)=...$?
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  5. #5
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    Re: Confirm if solutions are correct please.

    Quote Originally Posted by Siron View Post
    You can't say $\displaystyle \arcsin[\sin(x)]=x$, question b is an example of why it's not true (also think about it why it's not true in general).
    Actually I have given the answer to question b) with this quote:


    That means $\displaystyle \arcsin\left(\frac{-1}{2}\right)=...$?


    $\displaystyle -\frac{\pi}{6}$
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  6. #6
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    Re: Confirm if solutions are correct please.

    Just more one thing.

    sin(arcsin x ) = x if x belongs to interval [-1,1] right?

    because i wrote in interval [-pi/2, pi/2].
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: Confirm if solutions are correct please.

    Quote Originally Posted by Fabio010 View Post
    Just more one thing.

    sin(arcsin x ) = x if x belongs to interval [-1,1] right?

    because i wrote in interval [-pi/2, pi/2].
    Yes that's correct! Sorry, I hadn't seen that. Because the range of $\displaystyle y=\sin(x)$ is the closed interval [-1,1] that means the domain of (the inverse function) $\displaystyle y=\arcsin(x)$ is indeed the interval [-1,1].

    And indeed $\displaystyle \frac{-\pi}{6}$ is correct.
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  8. #8
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    Re: Confirm if solutions are correct please.

    Ok thanks for the help !
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  9. #9
    MHF Contributor Siron's Avatar
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    Re: Confirm if solutions are correct please.

    You're welcome!
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