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  1. #1
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    Proving...

    Q: Using the definitions of \mathrm{cosh} x and \mathrm{sinh} x in terms of e^x and e^{-x}, prove that \mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Air View Post
    Q: Using the definitions of \mathrm{cosh} x and \mathrm{sinh} x in terms of e^x and e^{-x}, prove that \mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1.

    Thanks in advance.
    Prove: \mathrm{cosh} 2x = 2 \mathrm{cosh}^2 x - 1.


    \cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 - 2}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1 .......
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Prove: \mathrm{cosh} 2x = 2 \mathrm{cosh}^2 x - 1.


    \cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 \mathbf{- 2}}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1 .......
    How did you get -2?
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  4. #4
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    Quote Originally Posted by Air View Post
    How did you get -2?
    (e^{x} + e^{-x})^2 = e^{2x} + 2 (e^x)(e^{-x}) + e^{-2x} = e^{2x} + 2 + e^{-2x}.

    Therefore e^{2x} + e^{-2x} = (e^{x} + e^{-x})^2 - 2.
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