Triple Angle Formulas/Tangent/Proof 2

Theorem

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

Proof

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.

$\blacksquare$