Multiple Angle Formula for Tangent

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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:

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

and so can be seen to hold.


$\map P 1$ is the case:

\(\ds \) \(\) \(\ds \frac {\ds \sum_{i \mathop = 0}^0 \paren {-1}^i \binom 1 {2 i + 1} \tan^{2 i + 1}\theta} {\ds \sum_{i \mathop = 0}^0 \paren {-1}^i \binom 1 {2 i} \tan^{2 i}\theta}\)
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^0 \dbinom 1 1 \tan^1 \theta} {\paren {-1}^0 \dbinom 1 0 \tan^0 \theta}\)
\(\ds \) \(=\) \(\ds \frac {\tan \theta} 1\)
\(\ds \) \(=\) \(\ds \tan \theta\)

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 $\map P {k - 2}$ 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:

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


For the second part:


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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:

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

$\blacksquare$


Also see

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