# Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean/Proof 1

## 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

 $\ds \sin n z$ $=$ $\ds \map \sin {\dfrac {n \pi} 2 + i n \ln \phi}$ $\ds$ $=$ $\ds \frac {e^{i \paren {\paren {n \pi / 2} + i n \ln \phi} } - e^{-i \paren {\paren {n \pi / 2} + i n \ln \phi} } } {2 i}$ Sine Exponential Formulation $\ds$ $=$ $\ds \frac {e^{i n \pi / 2} e^{-n \ln \phi} - e^{-i n \pi / 2} e^{n \ln \phi} } {2 i}$ $\ds$ $=$ $\ds \frac {e^{-n \ln \phi} \paren {\cos \frac {n \pi} 2 + i \sin \frac {n \pi} 2} - e^{n \ln \phi} \paren {\cos \paren {-\frac {n \pi} 2} + i \map \sin {-\frac {n \pi} 2} } } {2 i}$ Euler's Formula and Corollary $\ds$ $=$ $\ds \frac {e^{-n \ln \phi} \paren {i \sin \frac {n \pi} 2} - e^{n \ln \phi} \paren {i \map \sin {-\frac {n \pi} 2} } } {2 i}$ Cosine of Half-Integer Multiple of Pi $\ds$ $=$ $\ds \frac {i^n e^{-n \ln \phi} - \paren {-i}^n e^{n \ln \phi} } {2 i}$ Sine of Half-Integer Multiple of Pi and simplification $\ds$ $=$ $\ds \frac {i^{n - 1} \paren {e^{-n \ln \phi} + e^{n \ln \phi} } } 2$ simplification $\ds$ $=$ $\ds i^{n - 1} \frac {\phi^n + \frac 1 {\phi^n} } 2$ Exponential of Natural Logarithm $\ds$ $=$ $\ds i^{n - 1} \frac {\phi^n - \paren {-\frac 1 {\phi^n} } } 2$ $\ds$ $=$ $\ds i^{n - 1} \frac {\phi^n - \hat \phi^n} 2$ Reciprocal Form of One Minus Golden Mean

Setting $n = 1$:

$\sin z = i^0 \frac {\phi^1 - \hat \phi^1} 2 = \frac {\phi - \hat \phi} 2$

Thus:

 $\ds \dfrac {\sin n z} {\sin z}$ $=$ $\ds \dfrac {i^{n - 1} \frac {\phi^n - \hat \phi^n} 2} {\frac {\phi - \hat \phi} 2}$ $\ds$ $=$ $\ds i^{n - 1} \dfrac {\phi^n - \hat \phi^n } {\phi - \hat \phi}$ $\ds$ $=$ $\ds i^{n - 1} \dfrac {\frac {\phi^n - \hat \phi^n} {\sqrt 5} } {\frac {\phi - \hat \phi} {\sqrt 5} }$ $\ds$ $=$ $\ds i^{n - 1} \dfrac {F_n} {F_1}$ Euler-Binet Formula $\ds$ $=$ $\ds i^{n - 1} F_n$ Definition of Fibonacci Number: $F_1 = 1$

$\blacksquare$