Sine of Integer Multiple of Argument/Formulation 1

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Theorem

For $n \in \Z_{>0}$:

\(\ds \sin n \theta\) \(=\) \(\ds \sin \theta \paren {\paren {2 \cos \theta}^{n - 1} - \dbinom {n - 2} 1 \paren {2 \cos \theta}^{n - 3} + \dbinom {n - 3} 2 \paren {2 \cos \theta}^{n - 5} - \cdots}\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }\)


Proof

The proof proceeds by induction.


For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$\ds \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \sin \theta\) \(=\) \(\ds \sin \theta\)
\(\ds \) \(=\) \(\ds \sin \theta \binom {1 - 1 } 0 \paren {2 \cos \theta}^{1 - \paren {0 + 1} }\) Zero Choose Zero, Definition of Integer Power

So $\map P 1$ is seen to hold.


$\map P 2$ is the case:

\(\ds \sin 2 \theta\) \(=\) \(\ds 2 \sin \theta \cos \theta\) Double Angle Formula for Sine
\(\ds \) \(=\) \(\ds \sin \theta \binom {2 - 1} 0 \paren {2 \cos \theta}^{2 - \paren {0 + 1} }\)

So $\map P 2$ is also seen to hold.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true.


So this is our induction hypothesis:

$\ds \map \sin {n \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$

from which we are to show:

$\ds \map \sin {\paren {n + 1} \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n + 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k + 1} } }$


Induction Step

This is our induction step:

To proceed, we will require the following lemma:

Lemma

For $n \in \Z$:
\(\ds \map \cos {n \theta} \map \sin {\theta}\) \(=\) \(\ds \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta}\)

$\Box$


\(\ds \map \sin {\paren {n + 1} \theta}\) \(=\) \(\ds \map \sin {n \theta + \theta }\)
\(\ds \) \(=\) \(\ds \sin n \theta \cos \theta + \cos n \theta \sin \theta\) Sine of Sum
\(\ds \) \(=\) \(\ds \sin n \theta \cos \theta + \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta}\) Lemma above
\(\ds \) \(=\) \(\ds \sin n \theta \paren {2 \cos \theta } - \map \sin {\paren {n - 1 } \theta}\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } } \paren {2 \cos \theta } - \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n - 1 - \paren {2 k + 1} } }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k} } } - \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 2} } k \paren {2 \cos \theta}^{n - \paren {2 k + 2} } }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\paren {2 \cos \theta}^n + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k} } } + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - \paren {k - 1 + 2} } {k - 1 } \paren {2 \cos \theta}^{n - \paren {2 \paren {k - 1 } + 2} } } }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\paren {2 \cos \theta}^n + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \paren {\binom {n - \paren {k + 1} } k + \binom {n - \paren {k + 1} } {k - 1 } } \paren {2 \cos \theta}^{n - \paren {2 k} } } }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - k } k \paren {2 \cos \theta}^{n - \paren {2 k } } }\) Pascal's Rule
\(\ds \) \(=\) \(\ds \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - k } k \paren {2 \cos \theta}^{n - \paren {2 k } } }\)
\(\ds \) \(=\) \(\ds \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n + 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k + 1} } }\) adding and subtracting $1$


The result follows by the Principle of Mathematical Induction.

Therefore:

$\ds \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$

$\blacksquare$

Examples

Sine of Quintuple Angle

$\sin 5 \theta = \sin \theta \paren { \paren {2 \cos \theta}^4 - 3 \paren {2 \cos \theta}^2 + 1 }$


Sine of Sextuple Angle

$\sin 6 \theta = \sin \theta \paren { \paren {2 \cos \theta}^5 - 4 \paren {2 \cos \theta}^3 + 3 \paren {2 \cos \theta} }$


Sources

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