Sine of Integer Multiple of Argument/Formulation 8
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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:
| \(\ds \map \sin {n \theta}\) | \(=\) | \(\ds \paren {2 \cos \theta } \map \sin {\paren {n - 1 } \theta} - \map \sin {\paren {n - 2 } \theta}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \sin {\paren {n - 1 } \theta} \paren {\paren {2 \cos \theta } - \dfrac {\map \sin {\paren {n - 2 } \theta} } {\map \sin {\paren {n - 1 } \theta} } }\) | factoring out $\map \sin {\paren {n - 1 } \theta}$ |
Therefore $a_0 = 2 \cos \theta$
Once again, from Sine of Integer Multiple of Argument Formulation 4 we have:
| \(\ds \dfrac {\map \sin {n \theta} } {\map \sin {\paren {n - 1 } \theta} }\) | \(=\) | \(\ds 2 \cos \theta - \dfrac {\map \sin {\paren {n - 2 } \theta} } {\map \sin {\paren {n - 1 } \theta} }\) | dividing both sides by $\map \sin {\paren {n - 1 } \theta}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - 1 } \theta} } {\map \sin {\paren {n - 2 } \theta} } }\) | Move the numerator to the denominator |
In the equations above, let $n = n - k$:
| \(\ds \dfrac {\map \sin {\paren {n - k } \theta} } {\map \sin {\paren {n - k - 1 } \theta} }\) | \(=\) | \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - k - 1 } \theta} } {\map \sin {\paren {n - k - 2 } \theta} } }\) | ||||||||||||
| \(\ds \dfrac {\map \sin {\paren {n - k } \theta} } {\map \sin {\paren {n - \paren {k + 1} } \theta} }\) | \(=\) | \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - \paren {k + 1 } } \theta} } {\map \sin {\paren {n - \paren {k + 2 } } \theta} } }\) |
Therefore $a_1 = a_2 = \cdots = a_{n-3} = 2 \cos \theta$
Finally, let $k = n - 3$, then:
| \(\ds \dfrac {\map \sin {3 \theta} } {\map \sin {2 \theta } }\) | \(=\) | \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {2 \theta} } {\map \sin {\theta} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {2 \map \sin {\theta} \map \cos {\theta} } {\map \sin {\theta} } }\) | Double Angle Formula for Sine | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \cos \theta - \dfrac 1 {2 \map \cos \theta }\) |
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 $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.
$\blacksquare$
Examples
Sine of Quintuple Angle
- $\map \sin {5 \theta } = \map \sin {4 \theta} \paren { 2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2\cos \theta - \cfrac 1 {2 \cos \theta } } } }$
Sine of Sextuple Angle
- $\map \sin {6 \theta } = \map \sin {5 \theta} \paren { 2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2\cos \theta - \cfrac 1 {2\cos \theta - \cfrac 1 {2 \cos \theta} } } } }$