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Thread: Showing work for the natural number

  1. #1
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    Showing work for the natural number

    Given that
    $\displaystyle f(x) = \frac {e^x + e^{-x}}{2}$ and $\displaystyle g(x) = \frac {e^x - e^{-x}}{2}
    $ show that $\displaystyle
    [f(x)]^2 - [g(x)]^2 = 1$
    Ok here is what I figured out so far. By pluging those 2 functions in to the equation and squaring it I get
    $\displaystyle \frac {e^{2x} + e^{-2x}}{4} - \frac {e^{2x} + e^{-2x}}{4} = 1$.

    Here's my question though. Wouldn't that make it equal $\displaystyle 0$ once you subtract all the variables, as it would be $\displaystyle \frac {0}{4}$, which is equal to $\displaystyle 0$.
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  2. #2
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    $\displaystyle (\frac {e^{x} + e^{-x}}{2})^2\;=\; \frac {e^{2x} + e^{-2x} + 2}{4}$
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  3. #3
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    Quote Originally Posted by running-gag View Post
    $\displaystyle (\frac {e^{x} + e^{-x}}{2})^2\;=\; \frac {e^{2x} + e^{-2x} + 2}{4}$
    Thanks. I finally got it now.
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