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Cosh x / sinhx in the form of e^x

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    I was told that coshx = (e^x + e^-x) / 2 , why cosh2x = (e^2x - e^-2x) / 2 ?

    2. Relevant equations


    3. The attempt at a solution
     

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  2. jcsd
  3. Feb 13, 2016 #2

    LCKurtz

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    Because it is true for any value of x. Any number you can get with x you can get with 2x.
     
  4. Feb 13, 2016 #3
    If cosh3x then e^x is substituted with e^3x ??
     
  5. Feb 13, 2016 #4

    SammyS

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    Yes. Although you really should use parentheses for the expression making up the exponent unless it's written as a superscript.

    e^(3x) or e3x .
     
  6. Feb 14, 2016 #5
    Hi goldfish9776:

    I agree that the minus sign in
    cosh 2x = (e2x - e-2x) / 2 ​
    is peculiar.
    (e2x - e-2x) / 2 = sinh 2x.​

    Is the person who told you that
    cosh 2x = (e2x - e-2x) / 2 ​
    someone you would expect to be reliable?

    Regards,
    Buzz
     
  7. Feb 14, 2016 #6

    SammyS

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    @goldfish9776 ,

    You have a typo in the OP. The correct statement is:

    ##\displaystyle \cosh(2x)=\frac{e^{2x}+e^{-2x}}{2} ##
     
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