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

 * $\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - 1 } \theta} } {\map \sin {\paren {n - 2 } \theta} }} }$


 * $\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { 2 \cos \theta - \cfrac 1 {\cfrac {\paren {2 \cos \theta } \map \sin {\paren {n - 2 } \theta} - \map \sin {\paren {n - 3 } \theta}  } {\map \sin {\paren {n - 2 } \theta} }} }$


 * $\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { 2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac {\map \sin {\paren {n - 3 } \theta} } {\map \sin {\paren {n - 2 } \theta} }} }$

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