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Math Help - Hyperbolic functions

  1. #1
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    Hyperbolic functions

    Hi,

    I'm going through a past paper for an exam and found a problem which I don't know how to do, I am not sure if it is still examined any more but I am still interested in how you work it out:

    Using definitions \cosh(x) = \frac{1}{2}(e^x + e^{-x}) and \sinh(x) = \frac{1}{2}(e^x - e^{-x})

    \sinh(x - y) = \sinh(x)\cosh(y) - \cosh(x)\sinh(y)

    I think that it is related to double angle formula but I might be mistaken.
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  2. #2
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    Have you tried subbing in the definitions on each side?
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  3. #3
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    Well I stuck the terms in but they seem to cancel out each other so I assume I'm doing something wrong. Heres what I've done so far.

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

    I then did the same for the ones with the y and got.

    \frac{1}{4}((e^x - e^{-x})(e^y + e^{-y}) - (e^x + e^{-x})(e^y - e^{-y}))

    When i multiplied out i got:

    \frac{1}{4}((e^{xy} + e^{-xy} - e^{xy} - e^{xy}) - (e^{xy} - e^{-xy} + e^{-xy} - e^{xy}))

    and surely these terms all just cancel out, so where am I going wrong
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  4. #4
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    Quote Originally Posted by Beard View Post
    and surely these terms all just cancel out, so where am I going wrong
    all of them?

    Edit:

    Oh, also, your exponents aren't right

    Quote Originally Posted by Beard View Post
    \frac{1}{4}((e^x - e^{-x})(e^y + e^{-y}) - (e^x + e^{-x})(e^y - e^{-y}))
    Fine.

    Quote Originally Posted by Beard View Post
    When i multiplied out i got:

    \frac{1}{4}((e^{xy} + e^{-xy} - e^{xy} - e^{xy}) - (e^{xy} - e^{-xy} + e^{-xy} - e^{xy}))
    e^x * e^y = e^{xy} ?!
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  5. #5
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    Remember that e^x \cdot e^y=e^{x+y}, not e^{xy}.
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  6. #6
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    Sorry I haven't done maths in a looooong time
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  7. #7
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    I now get an alternative which I can sort of see is close to the answer but I'm not sure how to do the next step.

    \frac{1}{2}(e^{x - y} - e^{y - x})
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  8. #8
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    <br />
\frac{1}{2}(e^{x - y} - e^{y - x})<br />
= \frac{1}{2}(e^{x-y} - e^{-(x-y)})<br />
= \sinh (x-y)<br />
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