Multiple Angle Formula for Tangent

Theorem

 * $\ds \map \tan {n \theta} = \frac {\ds \sum_{i \mathop = 0}^{\floor{\frac {n - 1} 2} } \paren {-1}^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^{\floor {\frac n 2} } \paren {-1}^i \binom n {2 i} \tan^{2 i}\theta}$

Proof
Proof by induction:

For all $n \in \N_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds \map \tan {n \theta} = \frac {\ds \sum_{i \mathop = 0}^{\floor{\frac {n - 1} 2} } \paren {-1}^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^{\floor {\frac n 2} } \paren {-1}^i \binom n {2 i} \tan^{2 i}\theta}$

Basis for the Induction
$\map P 0$ is the case:

and so can be seen to hold.

$\map P 1$ is the case:

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 $\map P {k - 2}$ and $\map P {k - 1}$ is true, where $k > 2$ is an even number, then it logically follows that $\map P k$ and $\map P {k + 1}$ is true.

So this is our induction hypothesis:


 * $\ds \map \tan {\paren {k - 2} \theta} = \frac {\ds \sum_{i \mathop = 0}^{\frac k 2 - 2} \paren {-1}^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^{\frac k 2 - 1} \paren {-1}^i \binom {k - 2} {2 i} \tan^{2 i}\theta}$
 * $\ds \map \tan {\paren {k - 1} \theta} = \frac {\ds \sum_{i \mathop = 0}^{\frac k 2 - 1} \paren {-1}^i \binom {k - 1} {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^{\frac k 2 - 1} \paren {-1}^i \binom {k - 1} {2 i} \tan^{2 i}\theta}$

Then we need to show:


 * $\ds \map \tan {k \theta} = \frac {\ds \sum_{i \mathop = 0}^{\frac k 2 - 1} \paren {-1}^i \binom k {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^{\frac k 2} \paren {-1}^i \binom k {2 i} \tan^{2 i}\theta}$
 * $\ds \map \tan {\paren {k + 1} \theta} = \frac {\ds \sum_{i \mathop = 0}^{\frac k 2} \paren {-1}^i \binom {k + 1} {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^{\frac k 2} \paren {-1}^i \binom {k + 1} {2 i} \tan^{2 i}\theta}$

Induction Step
This is our induction step:

For the first part:

For the second part:

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

Therefore:
 * $\ds \forall n \in \N: \map \tan {n \theta} = \frac {\ds \sum_{i \mathop = 0}^{\floor{\frac {n - 1} 2} } \paren {-1}^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^{\floor {\frac n 2} } \paren {-1}^i \binom n {2 i} \tan^{2 i}\theta}$

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