Multiple Angle Formula for Tangent

Theorem

 * $\displaystyle \tan \left({n \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac {n - 1} 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac n 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i} \tan^{2 i}\theta}$

Proof
Proof by induction:

For all $n \in \N_{\ge0}$, let $P \left({b}\right)$ be the proposition:
 * $\displaystyle \tan \left({n \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac {n - 1} 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac n 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i} \tan^{2 i}\theta}$

Basis for the Induction
$P(0)$ is the case $\displaystyle \tan \left({0 \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{-1} \left({-1}\right)^i \binom 0 {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 0 {2 i} \tan^{2 i}\theta}$, which holds.

$P(1)$ is the case $\displaystyle \tan \left({\theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 1 {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 1 {2 i} \tan^{2 i}\theta}$, which also holds.

For $P(0)$:

For $P(1)$:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k-2}\right)$ and $P \left({k-1}\right)$ is true, where $k > 2$ is an even number, then it logically follows that $P \left({k}\right)$ and $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \tan \left({\left({k - 2}\right) \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta}$
 * $\displaystyle \tan \left({\left({k - 1}\right) \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 1} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 1} {2 i} \tan^{2 i}\theta}$

Then we need to show:
 * $\displaystyle \tan \left({k \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom k {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2} \left({-1}\right)^i \binom k {2 i} \tan^{2 i}\theta}$
 * $\displaystyle \tan \left({\left({k + 1}\right) \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2} \left({-1}\right)^i \binom {k + 1} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2} \left({-1}\right)^i \binom {k + 1} {2 i} \tan^{2 i}\theta}$

Induction Step
This is our induction step:

For the first part:

For the second part:

So $P\left({k-2}\right) \land P\left({k-1}\right) \implies P\left({k}\right) \land P\left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N: \tan \left({n \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac {n - 1} 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac n 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i} \tan^{2 i}\theta}$

Also see

 * $n = 2$: Double Angle Formula for Tangent


 * $n = 3$: Triple Angle Formula for Tangent


 * $n = 4$: Quadruple Angle Formula for Tangent


 * $n = 5$: Quintuple Angle Formula for Tangent