Quintuple Angle Formulas/Cosine/Proof 2
Jump to navigation
Jump to search
Theorem
- $\cos 5 \theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$
Proof
We have:
| \(\ds \cos 5 \theta + i \sin 5 \theta\) | \(=\) | \(\ds \paren {\cos \theta + i \sin \theta}^5\) | De Moivre's Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\cos \theta}^5 + \binom 5 1 \paren {\cos \theta}^4 \paren {i \sin \theta} + \binom 5 2 \paren {\cos \theta}^3 \paren {i \sin \theta}^2\) | Binomial Theorem | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \binom 5 3 \paren {\cos \theta}^2 \paren {i \sin \theta}^3 + \binom 5 4 \paren {\cos \theta} \paren {i \sin \theta}^4 + \paren {i \sin \theta}^5\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cos^5 \theta + 5 i \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta\) | substituting for binomial coefficients | |||||||||||
| \(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds 10 i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta\) | and using $i^2 = -1$ | ||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta\) | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds i \paren {5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta}\) | rearranging |
Hence:
| \(\ds \cos 5 \theta\) | \(=\) | \(\ds \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta\) | equating real parts in $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cos^5 \theta - 10 \cos^3 \theta \paren {1 - \cos^2 \theta} + 5 \cos \theta \paren {1 - \cos^2 \theta}^2\) | Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta\) | multiplying out and gathering terms |
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $21 \ \text{(a)}$