The imaginary unit with the golden ratio!

We have:

so:

As $sin(ix)=isinh(x)$ we can rewrite the expresion as $2isinh(ln\phi)$ And using the fact that $\phi$ is solution for $x^2-x-1=0$ we can find $i(\phi -\frac{1}{\phi})=i(\frac{\phi^2 -\phi -1 +\phi}{\phi})=i$

$2\sin (i \ln\phi) = 2i\sinh (\ln\phi)$

$2i\left(\frac{e^{\ln\phi} - e^{-\ln\phi}}{2}\right)$

$2i\left(\frac{\phi - \frac{1}{\phi}}{2}\right)$

$i\left(\phi - \frac{1}{\phi}\right)$

$i\left(\frac{1 + \sqrt{5}}{2} - \frac{2}{1 + \sqrt{5}}\right)$

$i\left(\frac{1 + \sqrt{5}}{2} - \frac{2(1 - \sqrt{5})}{-4}\right)$

$i\left(\frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2}\right)$

$i\left(\frac{2}{2}\right)$

$i$

$\large \boxed{2\sin (i \ln\phi) = i}$