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Proof using hyperbolic trig functions and complex variables

  1. Jul 9, 2015 #1
    1. Given, x + yi = tan^-1 ((exp(a + bi)). Prove that tan(2x) = -cos(b) / sinh(a)


    2. Relevant equations I have derived.
    tan(x + yi) = i*tan(x)*tanh(y) / 1 - i*tan(x)*tanh(y)

    tan(2x) = 2tanx / 1 - tan^2 (x)

    Exp(a+bi) = exp(a) *(cos(b) + i*sin(b))



    3. My attempt:
    By definition sinha = (exp(a) - exp (-a)) / 2

    Let x + yi = z

    tanz = exp(a + bi) = exp(a) * exp(bi)
    hence; exp(a) = tan(z) / exp(bi)
    exp(-a) = exp(bi) /tan(z)
    therefore; exp(a) - exp(-a) = (tan(z) / exp(bi) ) * ( exp(bi) / tan(z) )
    simplifying; exp(a) - exp(-a) = (tan^2(z) - exp(2bi)) / (exp(bi) * tan(z))
    hence, sinh (a) = (exp(a) - exp(-a))/ 2 = (tan^2(z) - exp(2bi)) / 2(exp(bi) * tan(z))

    Now am gonna find -cos(b) --------

    tanz = exp(a) * (cos(b) + isin(b))
    cos(b) = (tan(z) / exp(a)) - isin(b)
    -cos(b) = isin(b) - (tan(z) / exp(a))
    simplifying;
    -cos(b) = ( iexp(a)*sin(b) - tan(z) ) / exp(a)

    therefore;
    -cos(b) / sinh (a) =((( 2iexp(a + bi) * sin(b) tan(z) - 2exp(bi) * tan^2 (z) ))) / (( exp(a) tan^2(z) - exp(a +2bi) ))

    this is pretty much my best attempt...:'( I must be missing theory please help....I will really appreciate it!!!!
     
  2. jcsd
  3. Jul 9, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    You can factor out several things which makes the expression shorter.
    Also, the left side is real, so calculating the real part should be sufficient.
     
  4. Jul 9, 2015 #3
    Perhaps, but I'd seem a bit illogical to do that since they said the right hand side should be tan(2x)......sighz the problem is either rather difficult or am missing a simple piece of theory to make it a 5 line proof (often the case) .
     
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