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Math Help - Trig Identities

  1. #1
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    Trig Identities

    Anyone give me a leg up on verifying the following identity?

    sinh(x+y) = sinh x cosh y + cosh x sinh y
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  2. #2
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    Hello, kaiser0792!

    Use the definitions: . \begin{array}{ccc}\sinh z &=& \dfrac{e^z - e^{-z}}{2} \\ \\[-3mm]\cosh z &=& \dfrac{e^z + e^{-z}}{2} \end{array}


    Verify: . \sinh(x+y) \:=\: \sinh x \cosh y + \cosh x \sinh y

    The right side is:

    \sinh x\cosh y + \cosh x \sinh y

    . . . . =\;\frac{e^x-e^{-x}}{2}\cdot\frac{e^y+e^{-y}}{2} + \frac{e^x + e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2}

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

    . . . . =\; \frac{2e^{x+y} - 2e^{-(x+y)}}{4}

    . . . . =\;\frac{e^{x+y} - e^{-(x+y)}}{2}

    . . . . =\;\sinh(x+y)

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  3. #3
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    Thanks Soroban

    Excellent Soroban, I didn't think of it.
    Thank you
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