Is there any combinatorial proof of the formula for $\tan n\theta$ where $n\in \mathbb{N}$? Then proofs that I know are by Induction and using de Moivre's Formula but recently one of my friend asked me about a combinatorial proof of the identity. I thought about it for a while but couldn't come up with anything.

Instead of devising a direct combinatorial argument I tried to devise argument for proving the identities for $\cos n\theta$ and $\sin n\theta$ but in vain.

I think that probably no such proof is there but I am not sure about it and so any help regarding the proof of the identity will be helpful.


By formula for $\cos n\theta, \sin n\theta, \tan n\theta$ I meant the following identities respectively,

$$\cos n\theta=\displaystyle\sum_{i=0}^n \binom{n}{i}\cos^i\theta\sin^{n-i}\theta\cos\left(\dfrac{1}{2}(n-i)\pi\right)$$

$$\sin n\theta=\displaystyle\sum_{i=0}^n \binom{n}{i}\cos^i\theta\sin^{n-i}\theta\sin\left(\dfrac{1}{2}(n-i)\pi\right)$$

$$\tan n\theta=\left(\dfrac{\displaystyle\sum_{i=0}^n \binom{n}{i}\cos^i\theta\sin^{n-i}\theta\sin\left(\dfrac{1}{2}(n-i)\pi\right)}{\displaystyle\sum_{i=0}^n \binom{n}{i}\cos^i\theta\sin^{n-i}\theta\cos\left(\dfrac{1}{2}(n-i)\pi\right)}\right)$$

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    $\begingroup$ This is not directly related, but this paper has some examples of combinatorial proofs on trigonometric functions (not your identities, unfortunately.) $\endgroup$ – Eric Stucky Dec 12 '14 at 4:56
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    $\begingroup$ You're probably not still interested in this anymore, but I just happened across this combinatorial interpretation of the tangent function and I'm now wondering if there could be any hope of using it to get the desired proof. $\endgroup$ – Eric Stucky Oct 10 '15 at 6:56

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