Quadruple Angle Formulas/Tangent
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Theorem
- $\tan 4 \theta = \dfrac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}$
where $\tan$ denotes tangent.
Proof
\(\ds \tan 4 \theta\) | \(=\) | \(\ds \frac {\sin 4 \theta} {\cos 4 \theta}\) | Tangent is Sine divided by Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta} {8 \cos^4 \theta - 8 \cos^2 \theta + 1}\) | Quadruple Angle Formula for Sine and Quadruple Angle Formula for Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \frac {\tan \theta} {\cos^2 \theta} - 8 \tan^3 \theta} {8 - \frac 8 {\cos^2 \theta} + \frac 1 {\cos^4 \theta} }\) | dividing top and bottom by $\cos^4 \theta$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \tan \theta \sec^2 \theta - 8 \tan^3 \theta} {8 - 8 \sec^2 \theta + \sec^4 \theta}\) | Secant is Reciprocal of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \tan \theta \paren {1 + \tan^2 \theta} - 8 \tan^3 \theta} {8 - 8 \paren {1 + \tan^2 \theta} + \paren {1 + \tan^2 \theta}^2}\) | Difference of Squares of Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}\) | multiplying out and gathering terms |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.49$