Triple Angle Formulas/Sine/Proof 2

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Theorem

$\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$


Proof

We have:

\(\displaystyle \cos 3 \theta + i \sin 3 \theta\) \(=\) \(\displaystyle \paren {\cos \theta + i \sin \theta}^3\) De Moivre's Formula
\(\displaystyle \) \(=\) \(\displaystyle \paren {\cos \theta}^3 + \binom 3 1 \paren {\cos \theta}^2 \paren {i \sin \theta}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \binom 3 2 \paren {\cos \theta} \paren {i \sin \theta}^2 + \paren {i \sin \theta}^3\) Binomial Theorem
\(\displaystyle \) \(=\) \(\displaystyle \cos^3 \theta + 3 i \cos^2 \theta \sin \theta + 3 i^2 \cos \theta \sin^2 \theta + i^3 \sin^3 \theta\) substituting for binomial coefficients
\(\displaystyle \) \(=\) \(\displaystyle \cos^3 \theta + 3 i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta\) $i^2 = -1$
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \cos^3 \theta - 3 \cos \theta \sin^2 \theta\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle i \paren {3 \cos^2 \theta \sin \theta - \sin^3 \theta}\) rearranging


Hence:

\(\displaystyle \sin 3 \theta\) \(=\) \(\displaystyle 3 \cos^2 \theta \sin \theta - \sin^3 \theta\) equating imaginary parts in $(1)$
\(\displaystyle \) \(=\) \(\displaystyle 3 \paren {1 - \sin^2 \theta} \sin \theta - \sin^3 \theta\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 3 \sin \theta - 4 \sin^3 \theta\) multiplying out and gathering terms

$\blacksquare$