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