Quadruple Angle Formulas/Sine/Proof 2
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Theorem
- $\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$
Proof
We have:
\(\ds \cos 4 \theta + i \sin 4 \theta\) | \(=\) | \(\ds \paren {\cos \theta + i \sin \theta}^4\) | De Moivre's Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4\) | Binomial Theorem | ||||||||||
\(\ds \) | \(=\) | \(\ds \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta\) | substituting for binomial coefficients | |||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds 4 i \cos \theta \sin^3 \theta + \sin^4 \theta\) | and using $i^2 = -1$ | ||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta\) | rearranging |
Hence:
\(\ds \sin 4 \theta\) | \(=\) | \(\ds 4 \cos^3 \theta \sin \theta - 4 \cos \theta \sin^3 \theta\) | equating imaginary parts in $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 \cos \theta \paren {1 - \sin^2 \theta} \sin \theta - 4 \cos \theta \sin^3 \theta\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 \cos \theta \sin \theta - 8 \sin^3 \theta \cos \theta\) | multiplying out and gathering terms |
$\blacksquare$