Triple Angle Formulas/Tangent

Theorem

$\tan 3 \theta = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$

where $\tan$ denotes tangent.

Corollary 1

$\tan 3 \theta = \tan \theta \dfrac {4 \cos^2 \theta - 1} {4 \cos^2 \theta - 3}$

Proof 1

\(\ds \tan 3 \theta\)

\(=\)

\(\ds \frac {\sin 3 \theta} {\cos 3 \theta}\)

Tangent is Sine divided by Cosine

\(\ds \)

\(=\)

\(\ds \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}\)

Triple Angle Formula for Sine and Triple Angle Formula for Cosine

\(\ds \)

\(=\)

\(\ds \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}\frac {\cos^3 \theta} {\cos^3 \theta}\)

\(\ds \)

\(=\)

\(\ds \frac {\frac {3 \tan \theta} {\cos^2 \theta} - 4 \tan^3 \theta} {4 - \frac {3 \cos \theta} {\cos^3 \theta} }\)

Tangent is Sine divided by Cosine

\(\ds \)

\(=\)

\(\ds \frac {3 \tan \theta \sec^2 \theta - 4 \tan^3 \theta} {4 - 3 \sec^2 \theta}\)

Secant is Reciprocal of Cosine

\(\ds \)

\(=\)

\(\ds \frac {3 \tan \theta \paren {1 + \tan^2 \theta} - 4 \tan^3 \theta} {4 - 3 \paren {1 + \tan^2 \theta} }\)

Difference of Squares of Secant and Tangent

\(\ds \)

\(=\)

\(\ds \frac {3 \tan \theta + 3 \tan^3 \theta - 4 \tan^3 \theta} {4 - 3 - 3 \tan^2 \theta}\)

multiplying out

\(\ds \)

\(=\)

\(\ds \frac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}\)

gathering terms

$\blacksquare$

Proof 2

Let $\theta$ be such that $\tan 2 \theta$ is defined.

Then:

\(\ds \tan 3 \theta\)

\(=\)

\(\ds \dfrac {\tan \theta + \tan 2 \theta} {1 - \tan \theta \tan 2 \theta}\)

Tangent of Sum

\(\ds \)

\(=\)

\(\ds \dfrac {\tan \theta + \dfrac {2 \tan \theta} {1 - \tan^2 \theta} } {1 - \tan \theta \dfrac {2 \tan \theta} {1 - \tan^2 \theta} }\)

Double Angle Formula for Tangent

\(\ds \)

\(=\)

\(\ds \dfrac {\tan \theta \paren {1 - \tan^2 \theta} + 2 \tan \theta} {\paren {1 - \tan^2 \theta} - 2 \tan^2 \theta}\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac {\tan \theta - \tan^3 \theta + 2 \tan \theta} {1 - \tan^2 \theta - 2 \tan^2 \theta}\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}\)

simplifying

$\Box$

Now suppose $\theta$ is such that $\tan 2 \theta$ is not defined.

Then:

$2 \theta = \dfrac \pi 2 + n \pi$

for some integer $n$.

Hence:

$\theta = \dfrac \pi 4 + \dfrac {n \pi} 2$

For $n$ even we will then have:

$\tan \theta = 1$

and:

$\tan 3 \theta = -1$

For $n$ odd we will then have:

$\tan \theta = -1$

and:

$\tan 3 \theta = 1$

It is then directly verified that the Triple Angle Formula for Tangent holds for these special cases where $\tan 2 \theta$ is undefined.

$\blacksquare$

Proof 3

From Tangent of Sum of Three Angles:

$\map \tan {A + B + C} = \dfrac {\tan A + \tan B + \tan C - \tan A \tan B \tan C} {1 - \tan B \tan C - \tan C \tan A - \tan A \tan B}$

The result follows by setting $\theta = A = B = C$.

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.46$