Triple Angle Formulas/Sine/Proof 2
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Theorem
- $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$
Proof
We have:
| \(\ds \cos 3 \theta + i \sin 3 \theta\) | \(=\) | \(\ds \paren {\cos \theta + i \sin \theta}^3\) | De Moivre's Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\cos \theta}^3 + \binom 3 1 \paren {\cos \theta}^2 \paren {i \sin \theta}\) | ||||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \binom 3 2 \paren {\cos \theta} \paren {i \sin \theta}^2 + \paren {i \sin \theta}^3\) | Binomial Theorem | ||||||||||
| \(\ds \) | \(=\) | \(\ds \cos^3 \theta + 3 i \cos^2 \theta \sin \theta + 3 i^2 \cos \theta \sin^2 \theta + i^3 \sin^3 \theta\) | substituting for binomial coefficients | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cos^3 \theta + 3 i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta\) | $i^2 = -1$ | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \cos^3 \theta - 3 \cos \theta \sin^2 \theta\) | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds i \paren {3 \cos^2 \theta \sin \theta - \sin^3 \theta}\) | rearranging |
Hence:
| \(\ds \sin 3 \theta\) | \(=\) | \(\ds 3 \cos^2 \theta \sin \theta - \sin^3 \theta\) | equating imaginary parts in $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \paren {1 - \sin^2 \theta} \sin \theta - \sin^3 \theta\) | Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \sin \theta - 4 \sin^3 \theta\) | multiplying out and gathering terms |
$\blacksquare$