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
From Sine of Integer Multiple of Argument Formulation 4 we have:

Therefore $a_0 = 2 \cos \theta$

Once again, from Sine of Integer Multiple of Argument Formulation 4 we have:

In the equations above, let $n = n - k$:

Therefore $a_1 = a_2 = \cdots = a_{n-3} = 2 \cos \theta$

Finally, let $k = n - 3$, then:

Therefore $a_{n - 2} = 2 \cos \theta$

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