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Math Help - Noooo, I got wasted o.O

  1. #1
    Junior Member Lonehwolf's Avatar
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    Noooo, I got wasted o.O

    Prove teh Identitiy

    <br />
cosh (x+ y) = cosh x cosh y + sinh x sinh y<br />

    from the basic definitions of the hyperbolic functions. Hence, or otherwise show that if x and y satisfy the equations

    cosh x cosh y = 2,
    cosh x sinh y = -1

    then x = -y = +or- ln (1+*squareroot* of 2)

    I dunno how to use the math thingy too well so xD sorry if it's messed up.

    This question just has me staring, don't even know what I should be doing.
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,
    Quote Originally Posted by Lonehwolf View Post
    Prove teh Identitiy

    <br />
cosh (x+ y) = cosh x cosh y + sinh x sinh y<br />

    from the basic definitions of the hyperbolic functions.
    Use the "basic definitions" :
    \cosh(x)=\frac{e^x+e^{-x}}{2}
    \sinh(x)=\frac{e^x-e^{-x}}{2}

    Start from the right-hand side of the equality if it can help you.

    ~~~~~~~~~~~~~~~~~~~~
    also remember these rules :
    a^b a^c=a^{b+c}
    a^0=1
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    Hence, or otherwise show that if x and y satisfy the equations

    cosh x cosh y = 2,
    cosh x sinh y = -1

    then x = -y = +or- ln (1+*squareroot* of 2)

    I dunno how to use the math thingy too well so xD sorry if it's messed up.

    This question just has me staring, don't even know what I should be doing.
    shouldn't it be sinh(x)sinh(y)=-1 for the second equation ?

    If so, just add the two equation :
    \cosh(x)\cosh(y)+\sinh(x)\sinh(y)=1

    But the left-hand side of the equation is just \cosh(x+y)

    So 1=\cosh(x+y)=\frac{e^{x+y}+e^{-(x+y)}}{2}

    Let X=e^{x+y} :

    1=\frac{X+\frac 1X}{2}
    multiply both sides by 2X :
    2X=X^2+1

    X^2-2X+1=(X-1)^2=0

    hence X=e^{x+y}=1

    so \boxed{x+y=0}

    __________________________________
    You can see that \cosh(-x)=\cosh(x)

    Hence 2= \cosh(x)\cosh(y)=\cosh(x)\cosh(-x)=\left(\cosh(x)\right)^2

    So \cosh(x)=\pm \sqrt{2}

    Can you try to do this one ? There's just a quadratic to solve
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  4. #4
    Junior Member Lonehwolf's Avatar
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    Sorry for the late reply, ended up making a format and was left for three days without any access to internet >.<

    Either way, I'll use the first question's guidance you gave me to it, hope it ends up somewhere xD

    For the second one, yeh, it was sinh(x)sinh(y) = -1

    Pretty good of you to notice my... uhh... testing the community?

    But since I have a private lesson now I can't take any time trying those, it was from an Advanced Level past paper, and golly gosh you're good >_<
    Last edited by mr fantastic; November 28th 2008 at 03:33 AM. Reason: Replaced damn with golly gosh (lol)
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  5. #5
    Junior Member Lonehwolf's Avatar
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    Ok, got the time to check them out and I solved both successfully

    Thanks moo, tyvm
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