Quintuple Angle Formulas/Tangent
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Theorem
- $\tan 5 \theta = \dfrac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$
where $\tan$ denotes tangent.
Proof
\(\ds \tan 5 \theta\) | \(=\) | \(\ds \frac {\sin 5 \theta} {\cos 5 \theta}\) | Tangent is Sine divided by Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta} {16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta}\) | Quintuple Angle Formulas | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {5 \frac {\tan \theta} {\cos^4 \theta} - 20 \frac {\tan^3 \theta} {\cos^2 \theta} + 16 \tan^5 \theta} {16 \cos \theta - \frac {20} {\cos^2 \theta} + \frac 5 {\cos^4 \theta} }\) | dividing top and bottom by $\cos^5 \theta$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {5 \tan \theta \sec^4 \theta - 20 \tan^3 \theta \sec^2 \theta + 16 \tan^5 \theta} {16 - 20 \sec^2 \theta + 5 \sec^4 \theta}\) | Secant is Reciprocal of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {5 \tan \theta \paren {1 + \tan^2 \theta}^2 - 20 \tan^3 \theta \paren {1 + \tan^2 \theta} + 16 \tan^5 \theta} {16 - 20 \paren {1 + \tan^2 \theta} + 5 \paren {1 + \tan^2 \theta}^2}\) | Difference of Squares of Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}\) | multiplying out and gathering terms |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.52$