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Math Help - proving hyperbolic Identities

  1. #1
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    proving hyperbolic Identities

    Prove the identities:
    1.) \sinh(x+y) = \sinh{x}\cosh{y} + \cosh{x}\sinh{y}

    2.) \tanh(\ln{x}) = \frac{x^2 - 1}{x^2 + 1}

    thanks ahead
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Prove the identities:
    1.) \sinh(x+y) = \sinh{x}\cosh{y} + \cosh{x}\sinh{y}

    2.) \tanh(\ln{x}) = \frac{x^2 - 1}{x^2 + 1}

    thanks ahead
    1.) sinh(x+y) = sinh(x)cosh(y) +cosh(x)sin(y)

    I have done this before.
    I don't know how link, so here is a "copy-paste" of it:

    ----sinh u = (1/2)[e^u -e^(-u)]
    ----cosh u = (1/2)[e^u +e^(-u)]
    ----a^b * a^c = a^(b+c)

    sinh(x+y) = sinh x cosh y + cosh x sinh y

    We develop the RHS.
    RHS =
    = {(1/2)[e^x -e^(-x)] *(1/2)[e^y +e^(-y)]} +{(1/2)[e^x +e^(-x)] *(1/2)[e^y -e^(-y)]}

    = (1/4)[e^x -e^(-x)][e^y +e^(-y)] +(1/4)[e^x +e^(-x)][e^y -e^(-y)]

    Doing FOILs in the expansion,

    = (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y)] +(1/4)[e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

    = (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y) +e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

    = (1/4)[2e^(x+y) -2e^(-x-y)]

    = (1/2)[e^(x+y) -e^(-x-y)]

    = (1/2)[e^(x+y) -e^[-(x+y)]

    = sinh (x+y)

    = LHS

    Therefore, proven.

    -------------------------------------------
    2) tanh(ln(x)) = (x^2 -1) /(x^2 +1)

    tanh(ln(x)) =
    = sinh(ln(x)) /cosh(ln(x))
    = (e^ln(x) -e^(-ln(x)) / (e^ln(x) +e^(-ln(x))
    = (x -x^(-1)) / (x +x^(-1))
    = (x -1/x) / (x +1/x)
    = [(x^2 -1)/x] / [(x^2 +1)/x]
    = (x^2 -1) / (x^2 +1)

    Proven.

    Note:
    e^(-ln(x))
    = 1 /e^ln(x)
    = 1 /x
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  3. #3
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    Unhappy

    Quote Originally Posted by ticbol View Post
    1.) sinh(x+y) = sinh(x)cosh(y) +cosh(x)sin(y)

    I have done this before.
    I don't know how link, so here is a "copy-paste" of it:

    ----sinh u = (1/2)[e^u -e^(-u)]
    ----cosh u = (1/2)[e^u +e^(-u)]
    ----a^b * a^c = a^(b+c)

    sinh(x+y) = sinh x cosh y + cosh x sinh y

    We develop the RHS.
    RHS =
    = {(1/2)[e^x -e^(-x)] *(1/2)[e^y +e^(-y)]} +{(1/2)[e^x +e^(-x)] *(1/2)[e^y -e^(-y)]}

    = (1/4)[e^x -e^(-x)][e^y +e^(-y)] +(1/4)[e^x +e^(-x)][e^y -e^(-y)]

    Doing FOILs in the expansion,

    = (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y)] +(1/4)[e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

    = (1/4)[e^(x+y) +e^(x-y) -e^(-x+y) -e^(-x-y) +e^(x+y) -e^(x-y) +e^(-x+y) -e^(-x-y)]

    = (1/4)[2e^(x+y) -2e^(-x-y)]

    = (1/2)[e^(x+y) -e^(-x-y)]

    = (1/2)[e^(x+y) -e^[-(x+y)]

    = sinh (x+y)

    = LHS

    Therefore, proven.

    -------------------------------------------
    but its done backwards ...
    i followed from the end to the start
    but im still confused how
    = (1/2)[e^(x+y) -e^(-x-y)]
    became
    = (1/4)[2e^(x+y) -2e^(-x-y)]
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  4. #4
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    but its done backwards ...
    i followed from the end to the start
    but im still confused how
    = (1/2)[e^(x+y) -e^(-x-y)]
    became
    = (1/4)[2e^(x+y) -2e^(-x-y)]
    Because he factored out a two,
    (2/4)[e^(x+y)-e^(-x-y)]
    Simplify,
    (1/2)[e^(x+y)-e^(-x-y)]
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  5. #5
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    Hello, ^_^Engineer_Adam^_^ !

    This is ticbol's solution . . . in LaTeX.
    . . (Don't mind me . . . I love doing this!)


    Ticbol wrote:

    1)\;\sinh(x+y) \:= \:\sinh(x)\cosh(y) + \cosh(x)\sin(y)

    . . \sinh u \,= \,\frac{e^u -e^{-u}}{2}\qquad \cosh u \,= \,\frac{e^u + e^{-u}}{2} \qquad (a^b)(a^c)\,=\,a^{b+c}

    We have: . \sinh(x+y) \:= \:\sinh x \cosh y + \cosh x \sinh y

    We develop the RHS.

    \text{RHS }\; = \;\left(\frac{e^x -e^{-x}}{2}\right)\left(\frac{e^y +e^{-y}}{2}\right) + \left(\frac{e^x +e^{-x}}{2}\right)\left(\frac{e^y -e^{-y}}{2}\right)

    . . = \;\frac{(e^x -e^{-x})(e^y +e^{-y})}{4} + \frac{(e^x +e^{-x})(e^y -e^{-y})}{4}

    Doing FOILs in the expansion:

    . . = \;\frac{1}{4}\left[e^{x+y} +e^{x-y} -e^{-x+y} -e^{-x-y}\right] +\frac{1}{4}\left[e^{x+y} -e^{x-y} +e^{-x+y} -e^{-x-y}\right]

    . . = \;\frac{1}{4}\left[e^{x+y} +e^{x-y} -e^{-x+y} -e^{-x-y} +e^{x+y} -e^{x-y} +e^{-x+y} -e^{-x-y}\right]

    . . = \;\frac{1}{4}\left[2e^{x+y} -2e^{-x-y}\right] \;= \;\frac{1}{2}\left[e^{x+y} -e^{-x-y}\right] \;= \;\frac{e^{x+y} -e^{-(x+y)}}{2}

    . . = \;\sinh (x+y) \;= \;\text{LHS}

    Therefore, proven.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    2)\;\;\tanh(\ln x ) \;= \;\frac{x^2 -1}{x^2 +1}

    \tanh(\ln x)\;= \;\frac{\sinh(\ln x)}{\cosh(\ln x )}\;= \;\frac{\frac{e^{\ln x} -e^{-\ln x}}{2}} {\frac{e^{\ln x} + e^{-\ln x}}{2}} \;= \; \frac{e^{\ln x} -e^{-\ln x}}{e^{\ln x} + e^{-\ln x}} = \; \frac{e^{\ln x} -e^{\ln x^{-1}}}{e^{\ln x} + e^{\ln x^{-1}}} \;= \;\frac{x -x^{-1}}{x +x^{-1}}

    Multiply by \frac{x}{x}\!:\;\;\frac{x^2 - 1}{x^2 + 1}

    Therefore, proven.

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