Sine of Integer Multiple of Argument/Formulation 3

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:

$\ds \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1 } \frac {\cos k \theta} {\cos^k \theta}$

$\blacksquare$