Quick indices question

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  1. music lover's Avatar
    1 01-08-2012  09:54
    • Full Member
    • Posts: 146
    Quick indices question
    I need to prove that 1-tanh^2x=sech^2x. So we have

    1-(\frac{e^x-e^-^x}{e^x+e^-^x})^2. But then I'm a bit confused on how they simplify it. Down to this:

    \frac{(e^x+e^-^x)^2-(e^x-e^-^x)^2}{(e^x+e^-^x)^2}
  2. ReMos's Avatar
    2 01-08-2012  09:58
    • Full Member
    • Posts: 148
    Re: Quick indices question
    try expanding the top.

    Spoiler:
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    you could also use the identity cos^2(theta) - sin^2(theta) = 1
    Last edited by ReMos; 01-08-2012 at 09:59.
  3. notnek's Avatar
    3 01-08-2012  10:07
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: Quick indices question
    1=\displaystyle \frac{(e^x+e^{-x})^2}{(e^x+e^{-x})^2}

    Does that help?
  4. Intriguing Alias's Avatar
    4 01-08-2012  10:10
    • TSR Idol
    • Location: Yorkshire
    Re: Quick indices question
    (Original post by music lover)
    I need to prove that 1-tanh^2x=sech^2x. So we have

    1-(\frac{e^x-e^-^x}{e^x+e^-^x})^2. But then I'm a bit confused on how they simplify it. Down to this:

    \frac{(e^x+e^-^x)^2-(e^x-e^-^x)^2}{(e^x+e^-^x)^2}
    simplify the top by either expanding and collecting or using difference of two squares.
  5. music lover's Avatar
    5 01-08-2012  10:14
    • Full Member
    • Posts: 146
    Re: Quick indices question
    Got it. Thanks .
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