# 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 $P \left({n}\right)$ 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:

 $\displaystyle$  $\displaystyle \frac {\displaystyle \sum_{i \mathop = 0}^{-1} \left({-1}\right)^i \binom 0 {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 0 {2 i} \tan^{2 i}\theta}$ $\displaystyle$ $=$ $\displaystyle \frac 0 {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 0 {2 i} \tan^{2 i}\theta}$ Because the upper bound is smaller than the lower bound, this results in a vacuous summation and thus is zero. $\displaystyle$ $=$ $\displaystyle \frac 0 {\left({-1}\right)^0 \dbinom 0 0 \tan^0 \theta}$ $\displaystyle$ $=$ $\displaystyle \frac 0 1$ $\displaystyle$ $=$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle \tan \left({0 \theta}\right)$

and so can be seen to hold.

$P(1)$ is the case:

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

and so is also seen to hold.

These two cases together form the 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:

 $\displaystyle \tan \left({k \theta}\right)$ $=$ $\displaystyle \tan \left({\left({k - 2}\right) \theta + 2 \theta}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\tan \left({\left({k - 2}\right) \theta}\right) + \tan \left({2 \theta}\right)} {1 - \tan \left({\left({k - 2}\right) \theta}\right) \tan \left({2 \theta}\right)}$ Tangent of Sum $\displaystyle$ $=$ $\displaystyle \frac {\frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} + \tan \left({2 \theta}\right)} {1 - \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} \tan \left({2 \theta}\right)}$ By Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \frac {\frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} + \frac {\displaystyle 2 \tan \theta} {\displaystyle 1 - \tan^2 \theta} } {1 - \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} \frac {\displaystyle 2 \tan \theta} {\displaystyle 1 - \tan^2 \theta} }$ By Double Angle Formula for Tangent $\displaystyle$ $=$ $\displaystyle \frac {\left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta}\right) \left({\displaystyle 1 - \tan^2 \theta}\right) - \left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta}\right) \left({\displaystyle 2 \tan \theta}\right)} {\left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta}\right) \left({\displaystyle 1 - \tan^2 \theta}\right) - \left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta}\right) \left({\displaystyle 2 \tan \theta}\right)}$

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:

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

$\blacksquare$