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Math Help - Simplify cosh(arsinh4)

  1. #1
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    Simplify cosh(arsinh4)

    Simplify cosh(arsinh4)

    How would I got about simplifying this?

    Thanks
    Last edited by mr fantastic; May 25th 2009 at 01:02 AM. Reason: Added the rest of the question from the title.
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  2. #2
    MHF Contributor Amer's Avatar
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    Ok

    arsinh4 = ln(4+ 17^1/2 )
    cosh (arsinh4)

    cosh (ln (4 + 17^1/2 ) = 1/2(e^ln(4 + 17^1/2 ) + e^-ln(4 + 17^1/2 ))

    cosh (ln (4 + 17^1/2 ) = 1/2( 4+17^1/2 - 1/(4 + 17^1/2 )

    cosh (ln (4 + 17^1/2 ) = 1/2 ( ((4+17^1/2)^(2) -1))/(4 + 17^1/2)

    = ( 16 + 8(17)^1/2 + 17 -1 )/(4 + 17^1/2 )
    = ( 32 + 8(17)^1/2 )/(4+17^1/2)
    multiply with (-4+17^1/2)/(-4+17^1/2)


    = ( 32(17)^1/2 + 8(17) - 4(32) - 4*8(17)^1/2 )/(17 - 16 )
    = (136 - 128 )/1 =
    you continue
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  3. #3
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    perhaps simpler:

    cosh^2(x)- sinh^2(x)= 1 so cosh(x)= \sqrt{1+ sinh^2(x)}. With x= arcsinh(4), that is cosh(arcsinh(4))= \sqrt{1+ sinh^2(arcsinh(4))}= \sqrt{1+ 16}= \sqrt{17}
    Last edited by HallsofIvy; May 25th 2009 at 05:08 AM.
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  4. #4
    MHF Contributor Amer's Avatar
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    But my solution is 8
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  5. #5
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    Quote Originally Posted by Amer View Post
    But my solution is 8
    Sorry, but \sqrt{17} is the correct answer.

    Quote Originally Posted by Amer View Post
    Ok

    arsinh4 = ln(4+ 17^1/2 )
    cosh (arsinh4)

    cosh (ln (4 + 17^1/2 ) = 1/2(e^ln(4 + 17^1/2 ) + e^-ln(4 + 17^1/2 ))

    cosh (ln (4 + 17^1/2 ) = 1/2( 4+17^1/2 - 1/(4 + 17^1/2 ) Mr F says: That red minus should be a plus.

    cosh (ln (4 + 17^1/2 ) = 1/2 ( ((4+17^1/2)^(2) -1))/(4 + 17^1/2)

    = ( 16 + 8(17)^1/2 + 17 -1 )/(4 + 17^1/2 )
    = ( 32 + 8(17)^1/2 )/(4+17^1/2)
    multiply with (-4+17^1/2)/(-4+17^1/2)


    = ( 32(17)^1/2 + 8(17) - 4(32) - 4*8(17)^1/2 )/(17 - 16 )
    = (136 - 128 )/1 =
    you continue
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  6. #6
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    Hi, thanks for all the help everybody but I have a query with HallsOfIvy

    <br /> <br />
cosh(arcsinh(4))= \sqrt{1+ sinh^2(arcsinh(4))}= \sqrt{1+ 16}= \sqrt{17}<br />

    you have sinh^2 and arcsinh, but how does 1 arcsinh cancel with sinh^2, I thought It would only cancel with one of the sinh giving you

    <br /> <br />
cosh(arcsinh(4))= \sqrt{1+ sinh(4)}<br />

    Is that right?

    Thanks

    p.s. HallsOfIvy from physicsforums.com?
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  7. #7
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    You're misunderstanding the statement:

    \sinh^2 ( \text{arcsinh} (4)) \equiv [\sinh(\text{arcsinh}(4))]^2 \equiv 4^2.
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  8. #8
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    Quote Originally Posted by thomas49th View Post
    Hi, thanks for all the help everybody but I have a query with HallsOfIvy

    <br /> <br />
cosh(arcsinh(4))= \sqrt{1+ sinh^2(arcsinh(4))}= \sqrt{1+ 16}= \sqrt{17}<br />

    you have sinh^2 and arcsinh, but how does 1 arcsinh cancel with sinh^2, I thought It would only cancel with one of the sinh giving you

    <br /> <br />
cosh(arcsinh(4))= \sqrt{1+ sinh(4)}<br />

    Is that right?
    No, it is not right. " sinh^2(x) means sinh(x)*sinh(x), not sinh(sinh(x)).


    Thanks

    p.s. HallsOfIvy from physicsforums.com?
    I know that guy- he's a troublemaker!
    Last edited by mr fantastic; May 25th 2009 at 07:25 PM. Reason: Fixed the close quote tag.
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  9. #9
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    ahh yes. cool. I see

    Thanks
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