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Math Help - hyperbolic function identity

  1. #1
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    hyperbolic function identity

    By writing JA for in cot^2 +1=cosec^2 determine the corresponding hyperbolic identify
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by gracy View Post
    By writing JA for in cot^2 +1=cosec^2 determine the corresponding hyperbolic identify
    The trick you need here is to change every term containing sin^2 to -sinh^2
    (after converting it so it contains sin's and cos's only) in a trig identity to get
    the corresponding hyperbolic identity.

    So

    cot^2 +1=cosec^2

    becomes:

    sin^2/cos^2 + 1 = 1/sin^2.

    So the corresponding hyperbolic identity is:

    -sinh^2 / cosh^2 + 1 = -1/sinh^2

    or:

    -tanh^2 + 1 = -cosech^2

    or:

    tanh^2 -1 = cosech^2

    RonL
    Last edited by CaptainBlack; February 25th 2007 at 02:45 AM.
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  3. #3
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    Hello, gracy!

    Two questions: What is "JA"? .And do they really expect us to know
    . . conversions between circular and hyperbolic trig?


    By writing in JA for: cot²x + 1 .= .csc²x
    determine the corresponding hyperbolic identify.

    I would begin with the first identity of hyperbolics: .cosh²x - sinh²x .= .1

    . . . . . . . . . . . . . .cosh²x . .sinh²x . . . . . .1
    Divide by sinh²x: . --------- - -------- .= . --------
    . . . . . . . . . . . . . .sinh²x . . sinh²x . . . .sinh²x


    . . and we get: .coth²x - 1 .= .csch²x

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