Cosine of Integer Multiple of Argument/Formulation 8

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Theorem

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


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

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

A terminal $a_{n-1} = \cos \theta$ term.


Proof

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

\(\ds \map \cos {n \theta}\) \(=\) \(\ds \paren {2 \cos \theta } \map \cos {\paren {n - 1 } \theta} - \map \cos {\paren {n - 2 } \theta}\)
\(\ds \) \(=\) \(\ds \map \cos {\paren {n - 1 } \theta} \paren {\paren {2 \cos \theta } - \dfrac {\map \cos {\paren {n - 2 } \theta} } {\map \cos {\paren {n - 1 } \theta} } }\) factoring out $\map \cos {\paren {n - 1 } \theta}$

Therefore $a_0 = 2 \cos \theta$


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

\(\ds \dfrac {\map \cos {n \theta} } {\map \cos {\paren {n - 1 } \theta} }\) \(=\) \(\ds 2 \cos \theta - \dfrac {\map \cos {\paren {n - 2 } \theta} } {\map \cos {\paren {n - 1 } \theta} }\) dividing both sides by $\map \cos {\paren {n - 1 } \theta}$
\(\ds \) \(=\) \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\paren {n - 1 } \theta} } {\map \cos {\paren {n - 2 } \theta} } }\) Move the numerator to the denominator


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

\(\ds \dfrac {\map \cos {\paren {n - k } \theta} } {\map \cos {\paren {n - k - 1 } \theta} }\) \(=\) \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\paren {n - k - 1 } \theta} } {\map \cos {\paren {n - k - 2 } \theta} } }\)
\(\ds \dfrac {\map \cos {\paren {n - k } \theta} } {\map \cos {\paren {n - \paren {k + 1} } \theta} }\) \(=\) \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\paren {n - \paren {k + 1 } } \theta} } {\map \cos {\paren {n - \paren {k + 2 } } \theta} } }\)


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


Finally, let $k = n - 2$, then:

\(\ds \dfrac {\map \cos {2 \theta} } {\map \cos \theta }\) \(=\) \(\ds 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\theta} } {\map \cos {0} } }\)
\(\ds \) \(=\) \(\ds 2 \cos \theta - \dfrac 1 {\map \cos \theta }\)


Therefore $a_{n - 1} = \cos \theta$


Therefore:

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

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

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

A terminal $a_{n - 1} = \cos \theta$ term.

$\blacksquare$


Examples

Cosine of Quintuple Angle

$\map \cos {5 \theta } = \map \cos {4 \theta} \paren { 2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2\cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {\cos \theta } } }} }$


Cosine of Sextuple Angle

$\cos 6 \theta = \cos 5 \theta \paren {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {\cos \theta} } } } } }$