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Math Help - Trigonometric - hyperbolic

  1. #1
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    Trigonometric - hyperbolic

    Hi guys,
    just to make sure on the method i used on solving the identity attached.Remedial Action.doc

    I first expanded the squared trigonometric functions and tangent, then substituted with j(phi) - converted to cosh and sinh etc. ending with : 1-1-sinh^2\phi /cosh^2\phi =-sinh^2\phi /cosh^2\phi
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  2. #2
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    Quote Originally Posted by trojsi View Post
    Hi guys,
    just to make sure on the method i used on solving the identity attached.Remedial Action.doc

    I first expanded the squared trigonometric functions and tangent, then substituted with j(phi) - converted to cosh and sinh etc. ending with : 1-1-sinh^2\phi /cosh^2\phi =-sinh^2\phi /cosh^2\phi
    1. Sort out your brackets!

    2. \cos(2 i \phi)=\cosh(2\phi)

    CB
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  3. #3
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    Quote Originally Posted by trojsi View Post
    Hi guys,
    just to make sure on the method i used on solving the identity attached.Remedial Action.doc

    I first expanded the squared trigonometric functions and tangent, then substituted with j(phi) - converted to cosh and sinh etc. ending with : 1-1-sinh^2\phi /cosh^2\phi =-sinh^2\phi /cosh^2\phi
    Dear trojsi,

    You do not need to expand the trignometric functions. Replace \phi~with~i\phi and use, \cos(i\phi)=\cosh \phi~and~\tan(i\phi)=i\tanh \phi

    The correct identity is, -\tanh^2\phi=1-\dfrac{\cosh 2\phi}{\cosh^2\phi}

    Hope this helps you.
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  4. #4
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    Thanks a lot, I manage to get it with those identities.
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