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Math Help - Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

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    Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

    Im not quite sure what I need to do,

    I know

    sech^2x = 1/cosh(x)

    I know tanhx = sinhx/coshx

    I know that in trig 1+tan^2x would be sec^2x

    and I know it is 1-tanh^2x because of the (-) in the sinhx

    But, I dont know how to put this all together to prove it?
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    Re: Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

    Quote Originally Posted by tjnortham View Post
    Im not quite sure what I need to do,

    I know

    sech^2x = 1/cosh(x)----------should be sech(x)=1/cosh(x)

    I know tanhx = sinhx/coshx

    I know that in trig 1+tan^2x would be sec^2x

    and I know it is 1-tanh^2x because of the (-) in the sinhx

    But, I dont know how to put this all together to prove it?
    I hope you know the definitions of hyperbolic functions.

    \cosh(x)=\frac{e^{-x}+e^x}{2}\mbox{ and }\sinh(x)=\frac{e^{-x}-e^x}{2}

    1) Find \tanh(x) and hence 1-\tanh^{2}(x).

    2) Find \text{sech}^{2}(x)=\frac{1}{\cosh^{2}(x)}

    Compare the two results.
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  3. #3
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    Re: Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

    Quote Originally Posted by tjnortham View Post
    Im not quite sure what I need to do,

    I know

    sech^2x = 1/cosh(x)

    I know tanhx = sinhx/coshx

    I know that in trig 1+tan^2x would be sec^2x

    and I know it is 1-tanh^2x because of the (-) in the sinhx

    But, I dont know how to put this all together to prove it?
    \displaystyle \begin{align*} \textrm{sech}\,{(x)} &= \frac{1}{\cosh{(x)}} \\ &= \frac{2}{e^x + e^{-x}} \\ \textrm{sech}^2{(x)} &= \left(\frac{2}{e^x + e^{-x}}\right)^2 \\ &=\frac{4}{e^{2x} + 2 + e^{-2x}} \end{align*}

    \displaystyle \begin{align*} \tanh{(x)} &= \frac{\sinh{(x)}}{\cosh{(x)}} \\ &= \frac{\frac{e^{x} - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}} \\ &= \frac{e^x - e^{-x}}{e^x + e^{-x}} \\ \tanh^2{(x)} &= \left(\frac{e^x - e^{-x}}{e^x + e^{-x}}\right)^2 \\ &= \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} \\ 1 - \tanh^2{(x)} &= 1 - \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} \\ &= \frac{e^{2x} + 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} - \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} \\ &= \frac{4}{e^{2x} + 2 + e^{-2x}} \\ &= \textrm{sech}^2{x} \end{align*}
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    Re: Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

    Quote Originally Posted by tjnortham View Post
    Im not quite sure what I need to do,

    I know

    sech^2x = 1/cosh(x)

    I know tanhx = sinhx/coshx

    I know that in trig 1+tan^2x would be sec^2x

    and I know it is 1-tanh^2x because of the (-) in the sinhx

    But, I dont know how to put this all together to prove it?
    I haven't seen Osborne's rule used in any of the previous methods. At least, not directly. You know that 1+tan^2(x)=sec^2(x)
    Osborne's rule simply states that you can convert trigonometric identities to hyperbolic identities, except where there is a product of sines, when you have to change the sign. You know that: you reference it in the op.

    So, why not just prove 1+tan^2(x)=sec^2(x) and change the signs where appropriate?
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