Sine of Integer Multiple of Argument/Formulation 8

Theorem
For $n \in \Z_{>1}$:


 * $\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$

where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.

Proof
Therefore: For $n \in \Z_{>1}$:


 * $\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$

where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.