# sin (z) = cosh (2)

• Dec 22nd 2010, 04:52 PM
dwsmith
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 $\sin{z}=\cosh{2}\mbox{?}$
• Dec 22nd 2010, 08:14 PM
Prove It
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 $\displaystyle (x, y) = \left(\frac{\pi}{2}-2i, 0\right)$.
• Dec 22nd 2010, 08:56 PM
dwsmith
Quote:

Originally Posted by Prove It

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

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?
• Dec 22nd 2010, 10:51 PM
Chris L T521
Quote:

Originally Posted by dwsmith
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}
\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
\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.