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Math Help - [SOLVED] Help simplify this hyperbolic trig function

  1. #1
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    [SOLVED] Help simplify this hyperbolic trig function

    This should be really easy, but I'm not getting it for whatever reason

    \sqrt{sinh^2(t) - cosh^2(t) + 1^2}

    The answer should be \sqrt{2}cosh(t) But I don't understand how. There must be some simple identity that I'm missing.
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  2. #2
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    Quote Originally Posted by downthesun01 View Post
    This should be really easy, but I'm not getting it for whatever reason

    \sqrt{sinh^2(t) - cosh^2(t) + 1^2}

    The answer should be \sqrt{2}cosh(t) But I don't understand how. There must be some simple identity that I'm missing.
    Remember that \cosh^2{t} - \sinh^2{t} = 1...
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    Quote Originally Posted by Prove It View Post
    Remember that \cosh^2{t} - \sinh^2{t} = 1...
    Yeah, I realize that but you're going to have to dumb down your point a bit for me haha..
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    Quote Originally Posted by downthesun01 View Post
    Yeah, I realize that but you're giving to have to dumb down your point a bit for me haha..
    This means \sinh^2{t} = \cosh^2{t} - 1.

    Substitute it in and simplify...


    Upon further inspection, in order to obtain the answer of \sqrt{2}\cosh{t}, the original expression needs to be \sqrt{\sinh^2{t} \color{red}+\color{black}\cosh^2{t} + 1^2}
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    I figured out the problem. The original expression is
     \sqrt{sinh^2(t) + (-cosh(t))^2 + 1^2}

    Which is the same as  \sqrt{sinh^2(t) + cosh^2(t) + 1^2}
    So,
    \sqrt{2cosh^2(t)}
    <br />
\sqrt{2}cosh(t)
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