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  1. #1
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    Post hyperbolic functions2

    hi again!

    how would you solve these functions:

    2cosh2x - sinh2x = 2

    and prove

    tanh^{-1}(\frac{x^2-a^2}{x^2+a^2}) = In(\frac{x}{a})

    thanks
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  2. #2
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    Quote Originally Posted by dadon View Post
    hi again!

    how would you solve these functions:

    2cosh2x - sinh2x = 2
    You need some brackets here I think you mean:

    2[\cosh^2(x)-\sinh^2(x)]=2

    One way is if you know the "Hyperbolic Pythagoras's" theorem:

    \cosh^2(x)-\sinh^2(x)=1

    when it follows at once, but that would be to simple, so we must suppose
    that you are expected to expand the hyperbolic funtions in terms of
    exponentials. So:

    \sinh(x)=\frac{e^x-e^{-x}}{2},

    and:

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

    Then:

    \sinh^2(x)=\frac{e^{2x}-2+e^{-2x}}{4},

    and:

    \cosh^2(x)=\frac{e^{2x}+2+e^{-2x}}{4},

    So:

    \cosh^2(x)-\sinh^2(x)=\frac{e^{2x}+2+e^{-2x}}{4} - \frac{e^{2x}-2+e^{-2x}}{4}=1,

    and the required result follows on multiplying both sides by 2

    RonL
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  3. #3
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    sorry the question was express cosh2x and sinhx in exponential form and hence solve:

    2cosh2x - sinh2x = 2
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  4. #4
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    Quote Originally Posted by dadon View Post
    2cosh2x - sinh2x = 2
    I am wondering if the first one isn't an identity, but to solve the equation for x?

    2cosh(2x) - sinh(2x) = 2

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

    So the equation becomes:
    (e^{2x} + e^{-2x}) - \frac{1}{2}(e^{2x} - e^{-2x}) = 2

    \frac{1}{2}e^{2x} + \frac{3}{2}e^{-2x} = 2

    e^{2x} + 3e^{-2x} = 4 <-- Multiply both sides by e^{2x}

    e^{4x} + 3 = 4e^{2x}

    e^{4x} - 4e^{2x} + 3 = 0

    Now, for ease of notation, let y = e^{2x}. Then e^{4x} = y^2:

    So the equation is now:
    y^2 - 4y + 3 = 0

    (y - 3)(y - 1) = 0

    y = 3 or y = 1 <-- Put the x back: y = e^{2x}:

    e^{2x} = 3 or e^{2x} = 1

    2x = ln(3) or 2x = ln(1) = 0

    x = \frac{1}{2}ln(3) or x = 0

    So the solution set for x is 0, \, \frac{1}{2}ln(3)

    -Dan
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  5. #5
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    Hello, dadon!

    By the way, natural log is \ln\;(LN), not In.


    Prove: \text{tanh}^{-1}\left(\frac{x^2-a^2}{x^2+a^2}\right) \:= \:\ln\left(\frac{x}{a}\right)

    Let u \:=\:\text{tanh}^{-1}\left(\frac{x^2-a^2}{x^2+a^2}\right) \quad\Rightarrow\quad \text{tanh }u \:=\:\frac{x^2-a^2}{x^2+a^2}

    From the definition of \text{tanh }u, we have: . \frac{e^u - e^{-u}}{e^u+ e^{-u}} \:=\:\frac{x^2-a^2}{x^2+a^2}

    On the left side, multiply top and bottom by e^u\!:\;\;\frac{e^{2u} - 1}{e^{2u} + 1} \:=\:\frac{x^2-a^2}{x^2+a^2}


    Then: . . . . (x^2+a^2)(e^{2u} - 1) \:=\:(x^2-a^2)(e^{2u} + 1)

    . . . . (x^2 + a^2)e^{2u} - (x^2+a^2) \:=\:(x^2-a^2)e^{2u} + (x^2 - a^2)

    . . (x^2+a^2)e^{2u} - (x^2-a^2)e^{2u} \:=\:(x^2-a^2) + (x^2 + a^2)

    . . . . . . . . . . . . . . . 2a^2e^{2u} \:=\:2x^2

    . . . . . . . . . . . . . . . . . e^{2u} \:=\:\frac{x^2}{a^2} \:=\:\left(\frac{x}{a}\right)^2

    . . . . . . . . . . . . . . . . . 2u \:=\:\ln\left(\frac{x}{a}\right)^2\:=\:2\ln\left(\  frac{x}{a}\right)

    . . . . . . . . . . . . . . . . . . u \:=\:\ln\left(\frac{x}{a}\right)


    Therefore: . \text{tanh}^{-1}\left(\frac{x^2-a^2}{x^2+a^2}\right)\;=\;\ln\left(\frac{x}{a}\righ  t)

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