Sine of Integer Multiple of Argument/Formulation 9

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


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

where $a_0 = a_2 = a_4 = \ldots = 2 \sin \theta$ and $a_1 = a_3 = a_5 = \ldots = -2 \sin \theta$.

Proof

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


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


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


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


 * $\sin n \theta = \map \cos {\paren {n - 1 } \theta} \paren { 2 \sin \theta + \cfrac 1 {-2 \sin \theta + \cfrac 1 {\cfrac {\paren {2 \sin \theta } \map \cos {\paren {n - 3 } \theta} + \map \sin {\paren {n - 4 } \theta}  } {\map \cos {\paren {n - 3 } \theta} }}} }$

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


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

where $a_0 = a_2 = a_4 = \ldots = 2 \sin \theta$ and $a_1 = a_3 = a_5 = \ldots = -2 \sin \theta$.