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    Something easy I shouldn't be struggling with but can't seem to do:

    A^2 + B^2 - 2AB\cos(k)

    Use the substitution \cos(k) = \tanh(2\theta) and  A = \sinh\theta and  B = \cosh\theta to show that A^2 + B^2 - 2AB\cos(k) = \sin(k)

    I can get as far as showing A^2 + B^2 - 2AB\cos(k) = (A^2 + B^2)\sin^2k but then keep seeming to go around in circles.
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    what are A and B??
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     A = \sinh\theta B = \cosh\theta
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    (Original post by barnisaurusrex)
     A = \sinh\theta B = \cosh\theta
    sinh^2x+cosh^2x=cosh2x

    2sinhxcosh2x=sinh2x

    Using those would finish off the question easily
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    (Original post by barnisaurusrex)
     A = \sinh\theta B = \cosh\theta
    Didn't pass the step that you got to to arrive at the answer so if you show your steps, I can give you some guidance.

    Alternatively, if you want to start it fresh, play with the idea  sin(k) = \sqrt{1-tanh^2(2 \theta)}= sech(2 \theta)
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    (Original post by Clarity Incognito)
    Didn't pass the step that you got to to arrive at the answer so if you show your steps, I can give you some guidance.

    Alternatively, if you want to start it fresh, play with the idea  sin(k) = \sqrt{1-tanh^2(2 \theta)}= sech(2 \theta)
    Already tried that one, and the identities above. I'm trying to eliminate A and B here and everything I try gets rid of some A's and B's and adds a few more. Can you suggest an immediate step from the point I got to?


    edit: hold on, how incredible retarded of me.  A^2 + B^2 = cosh2\theta = 1/sech2\theta = 1/sink.

    sigh. Knew it was something simple. Cheers CI.
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    (Original post by barnisaurusrex)
    Already tried that one, and the identities above. I'm trying to eliminate A and B here and everything I try gets rid of some A's and B's and adds a few more. Can you suggest an immediate step from the point I got to?
    A^2+B^2=sinh^2\theta + cosh^2\theta=cosh2\theta

    2ABcosk=sinh2\theta tanh2\theta=\frac{sinh^2 2\theta}{cosh2\theta}

    A^2+B^2-2ABcosk=cosh2\theta-\frac{sinh^2 2\theta}{cosh2\theta}=\frac{cosh  ^2 2\theta-sinh^2 2\theta}{cosh2\theta}=sech 2\theta=sink
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    (Original post by barnisaurusrex)
    Already tried that one, and the identities above. I'm trying to eliminate A and B here and everything I try gets rid of some A's and B's and adds a few more. Can you suggest an immediate step from the point I got to?


    edit: hold on, how incredible retarded of me.  A^2 + B^2 = cosh2\theta = 1/sech2\theta = 1/sink.

    sigh. Knew it was something simple. Cheers CI.
    No worries
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    I don't suppose you also get  2B^2 - 2ABcos(k) also equalling \sin(k) do you? My equation would work out nicely if it did but I can't seem to show it yet again, if it does. I have a feeling it doesn't though.
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    (Original post by barnisaurusrex)
    I don't suppose you also get  2B^2 - 2ABcos(k) also equalling \sin(k) do you? My equation would work out nicely if it did but I can't seem to show it yet again, if it does. I have a feeling it doesn't though.
    If  2B^2 - 2ABcos(k) = sin(k)

    then

     A^2 + B^2 - 2ABcos(k) = 2B^2 - 2ABcos(k)

    implies that

     A^2 = B^2

     sinh^2 \theta = cosh^2 \theta of which no solutions exist therefore

    sinh^2 \theta \not= cosh^2 \theta \forall \theta and so what we wanted to show is the same as sin(k) is not true.
 
 
 
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