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

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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

\(\ds \sin n z\) \(=\) \(\ds \map \sin {\dfrac {n \pi} 2 + i n \ln \phi}\)
\(\ds \) \(=\) \(\ds \frac {e^{i \paren {\paren {n \pi / 2} + i n \ln \phi} } - e^{-i \paren {\paren {n \pi / 2} + i n \ln \phi} } } {2 i}\) Euler's Sine Identity
\(\ds \) \(=\) \(\ds \frac {e^{i n \pi / 2} e^{-n \ln \phi} - e^{-i n \pi / 2} e^{n \ln \phi} } {2 i}\)
\(\ds \) \(=\) \(\ds \frac {e^{-n \ln \phi} \paren {\cos \frac {n \pi} 2 + i \sin \frac {n \pi} 2} - e^{n \ln \phi} \paren {\cos \paren {-\frac {n \pi} 2} + i \map \sin {-\frac {n \pi} 2} } } {2 i}\) 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: $F_1 = 1$

$\blacksquare$