Quintuple Angle Formulas/Tangent

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: $5.52$