Sine of Integer Multiple of Argument/Formulation 3
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Theorem
For $n \in \Z_{>0}$:
\(\ds \sin n \theta\) | \(=\) | \(\ds \sin \theta \cos^{n - 1} \theta \paren {1 + 1 + \frac {\cos 2 \theta} {\cos^2 \theta} + \frac {\cos 3 \theta} {\cos^3 \theta} + \cdots + \frac {\cos \paren {n - 1} \theta} {\cos^{n - 1} \theta} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}\) |
Proof
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
- $\ds \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop \ge 0} \frac {\cos k \theta} {\cos^k \theta}$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \sin \theta\) | \(=\) | \(\ds \sin \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta \cos^{1 - 1} \theta \paren 1\) |
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta \cos^{2 - 1} \theta \paren {1 + 1}\) |
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:
- $\ds \map \sin {n \theta} = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta } {\cos^k \theta}$
from which we are to show:
- $\ds \map \sin {\paren {n + 1} \theta} = \sin \theta \cos^n \theta \sum_{k \mathop = 0}^n \frac {\cos k \theta} {\cos^k \theta}$
Induction Step
This is our induction step:
For the first part:
\(\ds \map \sin {\paren {n + 1} \theta}\) | \(=\) | \(\ds \map \sin {n \theta + \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sin n \theta \cos \theta + \cos n \theta \sin \theta\) | Sine of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} } \cos \theta + \cos n \theta \sin \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta \paren {\cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} \cos \theta + \cos n \theta}\) | Factor out $\sin \theta$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} + \cos n \theta}\) | Definition of Integer Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} + \cos n \theta \frac {\cos^n \theta} {\cos^n \theta} }\) | multiply by $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^n \frac {\cos k \theta} {\cos^k \theta} }\) |
The result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1 } \frac {\cos k \theta} {\cos^k \theta}$
$\blacksquare$