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Math Help - inverse sin etc.

  1. #1
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    inverse sin etc.

    Ok i solved these problems.

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

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

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

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

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

    c) tg(arctg(-1))

    -1 belongs to /Re

    so solution is -1
    Last edited by Fabio010; November 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 \sin(\arcsin(x))=x and \tan[\arctan(x)]=x (if you know why)
    For b) first of all \sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=\frac{-1}{2}
    So that means \arcsin\left(\frac{-1}{2}\right) \in \[\frac{-\pi}{2},\frac{\pi}{2}\]

    So that means \frac{7\pi}{6} can't never be an answer because it's not in \[\frac{-\pi}{2},\frac{\pi}{2}\]
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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 \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
    \sin\left(\frac{7\pi}{6}\right)=-\sin\left(\frac{\pi}{6}\right)=\frac{-1}{2}
    That means \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 \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 \arcsin\left(\frac{-1}{2}\right)=...?


    -\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 y=\sin(x) is the closed interval [-1,1] that means the domain of (the inverse function) y=\arcsin(x) is indeed the interval [-1,1].

    And indeed \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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