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Math Help - Can't Find The Identity

  1. #1
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    Can't Find The Identity

    Anyone know of an identity for:

    cosh(x)sin(y)-sinh(x)cos(y)

    I know how this works if the hyperbolic cosine and sine were regular cosine and sine, but I cannot seem to find any identities online that relate these.

    Thanks in advance for the help.
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  2. #2
    Junior Member bleesdan's Avatar
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    Hyperbolic Functions -- from Wolfram MathWorld

    This has all the identities on it.

    It looks like an imaginary identity, but there's no i.

    Hold on

    OK, I got nothin'.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rman144 View Post
    Anyone know of an identity for:

    cosh(x)sin(y)-sinh(x)cos(y)

    I know how this works if the hyperbolic cosine and sine were regular cosine and sine, but I cannot seem to find any identities online that relate these.

    Thanks in advance for the help.
    I'm with bleesdan. I'm suspecting a typo?

    -Dan
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  4. #4
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    Reply

    Sorry guys, no typo. There is no imaginary part there. This is why the problem is proving to be such an issue. Anyone else have an idea?
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  5. #5
    Eater of Worlds
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    Here is one, for what it's worth:

    \frac{\sqrt{2}\left(sin(y+\frac{\pi}{4})-e^{2x}cos(y+\frac{\pi}{4})\right)}{2e^{x}}
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  6. #6
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    Thanks

    Thanks a lot. May I ask where you found that/ solved for that?
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  7. #7
    Eater of Worlds
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    I derived it.
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  8. #8
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    Hello, rman144!

    There is no identity for a "mixed" statement like this.
    . . We must derive one like galactus did.


    \cosh(x)\sin(y)-\sinh(x)\cos(y)
    Write all of them in exponential form . . .

    \frac{e^x+e^{-x}}{2}\cdot\frac{e^{iy} - e^{-iy}}{2i} \;- \;\frac{e^x-e^{-x}}{2}\cdot\frac{e^{iy} + e^{-iy}}{2}

    . . = \;\;\frac{e^{x+iy} - e^{x-iy} + e^{-x+iy} - e^{-x-iy}}{4i} \;- \;\frac{e^{x+iy} + e^{x-iy} - e^{-x+iy} - e^{-x-iy}}{4}


    Multiply the first fraction by \frac{-i}{-i}

    \frac{-ie^{x+iy} + ie^{x-iy} - ie^{-x+iy} + ie^{-x-iy}}{4} \;- \;\frac{e^{x+iy} + e^{x-iy} - e^{-x+iy} - e^{-x-iy}}{4}

    . . = \;\;\frac{1}{4}\bigg[-e^{x+iy} - ie^{x+iy} - e^{x-iy} + ie^{x-iy} + e^{-x+iy} - ie^{-x+iy} + e^{-x-iy} + ie^{-x-iy}\bigg]

    . . = \;\;\frac{1}{4}\bigg[-(1+i)\left(e^{x+iy} - e^{-(x+iy)}\right) \;- \;(1-i)\left(e^{x-iy} + e^{-(x-iy)}\right)\bigg]

    . . =\;\;-\frac{1}{2}\bigg[(1+i)\frac{e^{x+iy} - e^{-(x+iy)}}{2} \;+ \;(1-i)\frac{e^{x-iy} + e^{-(x-iy)}} {2}\bigg]

    . . = \;\;-\frac{1}{2}\bigg[(1+i)\sinh(x+iy) \;+ \;(1-i)\cosh(x-iy)\bigg]

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  9. #9
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    Thanks

    You guys are life savers.
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