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Math Help - sin (z) = cosh (2)

  1. #1
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    sin (z) = cosh (2)

    The question says to solve sin (z) = cosh (2) by equating the real and imaginary parts.

    \displaystyle\sin{z}=\sin{x}\cosh{y}+\mathbf{i}\co  s{x}\sinh{y} \ \ \mbox{and} \ \ \cosh{2}=\frac{e^2+e^{-2}}{2}

    These are the two equations to solve.

    \displaystyle \sin{x}\cosh{y}=\frac{e^2+e^{-2}}{2}\Rightarrow \ \ x=\frac{\pi}{2}+\pi k, \ k\in\mathbb{Z}, \ \mbox{and} \ y=2

    \displaystyle \cos{x}\sinh{y}=0\Rightarrow \ \ x=\frac{\pi}{2}+\pi k, \ k\in\mathbb{Z}, \ \mbox{or} \ y=0

    From this, I am just supposed to include when

    \displaystyle x=\frac{\pi}{2}+\pi k, \ k\in\mathbb{Z}, \ \mbox{and} \ y=2,

    then  \sin{z}=\cosh{2}\mbox{?}
    Last edited by dwsmith; December 22nd 2010 at 04:54 PM. Reason: Changed equation to equating
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  2. #2
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    From Equation 2, you have gotten two cases, \displaystyle x = \frac{\pi}{2} + \pi k, k \in \mathbf{Z} OR \displaystyle y = 2.


    Case 1: \displaystyle x = \frac{\pi}{2} + \pi k, k \in \mathbf{Z}.

    Substitute into Equation 1 to give

    \displaystyle \sin{\left(\frac{\pi}{2} + \pi k\right)}\cosh{y} = \cosh{2}

    \displaystyle (-1)^k\cosh{y} = \cosh{2}

    \displaystyle \cosh{y} = \frac{\cosh{2}}{(-1)^k}.

    But since \displaystyle \cosh{y} > 0 for all \displaystyle y, that means we can only accept even values of \displaystyle k, and therefore, that restricts \displaystyle x to be \displaystyle \frac{\pi}{2} + 2\pi n, n \in \mathbf{Z}.

    When that is the case, \displaystyle \cosh{y} = \cosh{2} \implies y = 2.

    So the first solution is \displaystyle (x,y) = \left(\frac{\pi}{2} + 2\pi n, 2\right), n \in \mathbf{Z}.


    Case 2: \displaystyle y = 0.

    Substitute into Equation 1 to give

    \displaystyle \sin{x}\cosh{0} = \cosh{2}

    \displaystyle \sin{x} = \cosh{2}

    \displaystyle x = \arcsin{(\cosh{2})}

    \displaystyle x = \frac{\pi}{2} - 2i according to Mathematica.

    So the second solution is \displaystyle (x, y) = \left(\frac{\pi}{2}-2i, 0\right).
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    Quote Originally Posted by Prove It View Post

    \displaystyle x = \arcsin{(\cosh{2})}
    How can this be solved by hand?

    I remember doing sin(arcCos) and other basic functions but is there a method for inverse trig, hyperbolic, and inverse hyperbolic equations?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dwsmith View Post
    How can this be solved by hand?

    I remember doing sin(arcCos) and other basic functions but is there a method for inverse trig, hyperbolic, and inverse hyperbolic equations?
    Recall that \arcsin(z) = -i\ln\left(iz+\sqrt{1-z^2}\right) for z\in\mathbb{C}.

    Since \cosh 2 \in \mathbb{C} we can use this formula.

    So, we see that

    \begin{aligned}<br />
   \arcsin(\cosh 2) &= -i\ln\left(i\cosh 2+\sqrt{1-\cosh^22}\right)\\\ &= -i\ln\left(i\cosh 2 + i\sinh 2\right)\\ &= -i\ln\left(ie^2\right)\\ &= -i\ln(i) -2i<br />
\end{aligned}

    Now recall that i=e^{i\frac{\pi}{2}}. Therefore, you end up with -i\ln\left(e^{i\frac{\pi}{2}}\right)-2i=\frac{\pi}{2}-2i, as given by Mathematica.
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