# Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean/Proof 2

## Theorem

Let $z = \dfrac \pi 2 + i \ln \phi$.

Then:

$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$

where:

$\phi$ denotes the golden mean
$F_n$ denotes the $n$th Fibonacci number.

## Proof

 $\ds \cos z$ $=$ $\ds \cos \left({\dfrac {\pi} 2 + i \ln \phi}\right)$ $\ds$ $=$ $\ds \frac {e^{i \left({\left({\pi / 2}\right) + i \ln \phi}\right)} + e^{-i \left({\left({\pi / 2}\right) + i \ln \phi}\right)} } 2$ Cosine Exponential Formulation $\ds$ $=$ $\ds \frac {e^{i \pi / 2} e^{-\ln \phi} + e^{-i \pi / 2} e^{\ln \phi} } 2$ $\ds$ $=$ $\ds \frac {e^{-\ln \phi} \left({\cos \frac \pi 2 + i \sin \frac \pi 2}\right) + e^{\ln \phi} \left({\cos \left({-\frac \pi 2}\right) + i \sin \left({-\frac \pi 2}\right)}\right)} 2$ Euler's Formula and Corollary $\ds$ $=$ $\ds \frac {e^{-\ln \phi} \left({i \sin \frac \pi 2}\right) + e^{\ln \phi} \left({i \sin \left({-\frac \pi 2}\right)}\right)} 2$ Cosine of Half-Integer Multiple of Pi $\ds$ $=$ $\ds \frac {i e^{-\ln \phi} - i e^{\ln \phi} } 2$ Sine of Half-Integer Multiple of Pi and simplification $\ds$ $=$ $\ds -i \frac {\phi - \frac 1 {\phi} } 2$ Exponential of Natural Logarithm $\ds$ $=$ $\ds -i \frac {\phi^2 - 1} {2 \phi}$ $\ds$ $=$ $\ds -i \frac {\phi} {2 \phi}$ Square of Golden Mean equals One plus Golden Mean $\ds$ $=$ $\ds \frac {-i} 2$

Then:

 $\ds \sin \left({n + 1}\right) z + \sin \left({n - 1}\right) z$ $=$ $\ds 2 \sin n z \cos z$ Simpson's Formula for Sine by Cosine $\ds$ $=$ $\ds -i \sin n z$