Sine of Integer Multiple of Argument/Formulation 4

Theorem

For $n \in \Z$:

\(\ds \map \sin {n \theta}\)

\(=\)

\(\ds \paren {2 \cos \theta } \map \sin {\paren {n - 1 } \theta} - \map \sin {\paren {n - 2 } \theta}\)

Proof

To proceed, we will require the following lemma:

Lemma

For $n \in \Z$:

\(\ds \map \cos {n \theta} \map \sin {\theta}\)

\(=\)

\(\ds \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta}\)

$\Box$

\(\ds \map \sin {n \theta}\)

\(=\)

\(\ds \map \sin {\paren {n - 1 } \theta + \theta }\)

\(\ds \)

\(=\)

\(\ds \map \sin {\paren {n - 1 } \theta} \cos \theta + \map \cos {\paren {n - 1 } \theta} \sin \theta\)

Sine of Sum

\(\ds \)

\(=\)

\(\ds \map \sin {\paren {n - 1 } \theta} \cos \theta + \map \sin {\paren {n - 1 } \theta} \map \cos {\theta} - \map \sin {\paren {n - 2 } \theta}\)

Lemma above

\(\ds \)

\(=\)

\(\ds \paren {2 \cos \theta } \map \sin {\paren {n - 1 } \theta} - \map \sin {\paren {n - 2 } \theta}\)

$\blacksquare$