MHB Finite Tangent product / quotient

AI Thread Summary
The discussion focuses on proving the identity involving the tangent function for real numbers and odd integers. It establishes that the ratio of tangent functions can be expressed as a product of specific tangent terms derived from the roots of the equation $\tan(m\theta) = \tan(mz)$. The proof utilizes the formula for $\tan(m\theta)$, which is based on de Moivre's theorem, and involves manipulating the roots to arrive at the desired equation. Participants acknowledge the elegance of the approaches used, particularly in relation to complex numbers. Overall, the thread emphasizes the mathematical derivation and its implications in trigonometric identities.
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Just for fun, eh...? (Heidy)For $$z \in \mathbb{R}$$, and $$m \in 2\mathbb{N}+1$$, show that:$$\frac{\tan mz}{\tan z}=\prod_{j=1}^{ \lfloor m/2 \rfloor } \tan\left(\frac{j\pi}{m}+z\right) \tan\left(\frac{j\pi}{m}-z\right) $$
 
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[sp]The numbers $\theta = \frac{j\pi}{m}+z \ (-\lfloor m/2 \rfloor \leqslant j \leqslant \lfloor m/2 \rfloor)$ are the roots of the equation $\tan(m\theta) = \tan(mz).$ The formula for $\tan(m\theta)$ (for an odd number $m$) in terms of $t = \tan\theta$ is $$\tan(m\theta) = \frac{{m\choose1}t - {m\choose3}t^3 + {m\choose5}t^5 - \ldots + (-1)^{\lfloor m/2 \rfloor}t^m} {1 - {m\choose2}t^2 + {m\choose4}t^4 - \ldots + (-1)^{\lfloor m/2 \rfloor}{m\choose m-1}t^{m-1}}.$$ So the equation obtained by putting that expression equal to $\tan(mz)$ has roots $\tan\left(\frac{j\pi}{m}+z\right) \ (-\lfloor m/2 \rfloor \leqslant j \leqslant \lfloor m/2 \rfloor).$ Multiply out the fraction and the equation becomes $(-1)^{\lfloor m/2 \rfloor}t^m + \ldots - \tan(mz) = 0.$ The product of the roots is the constant term divided by the coefficient of $t^m.$ Therefore $$\prod_{j=-\lfloor m/2 \rfloor}^{\lfloor m/2 \rfloor} \tan\left(\frac{j\pi}{m}+z\right) = (-1)^{\lfloor m/2 \rfloor}\tan(mz).$$ Now divide by the middle term of the product and pair off the remaining factors to get $$ (-1)^{\lfloor m/2 \rfloor}\frac{\tan(mz)}{\tan z} = \prod_{j=1}^{ \lfloor m/2 \rfloor } \tan\left(\frac{j\pi}{m}+z\right) \tan\left(z - \frac{j\pi}{m}\right).$$ Finally, change the sign of the second of each of those pairs of factors. That will introduce $\lfloor m/2 \rfloor$ changes of sign, which will cancel with those on the left side of the equation and result in $$\frac{\tan mz}{\tan z}=\prod_{j=1}^{ \lfloor m/2 \rfloor } \tan\left(\frac{j\pi}{m}+z\right) \tan\left(\frac{j\pi}{m}-z\right).$$[/sp]
 
@ Opalg...

I used the standard definition of the tangent function in complex terms,

$$\tan \theta = \frac{e^{i\theta}- e^{-i\theta}}{ e^{i\theta}+ e^{-i\theta}}$$But your approach is far more elegant. Very nicely doen indeed! (Yes)
 
DreamWeaver said:
@ Opalg...

I used the standard definition of the tangent function in complex terms,

$$\tan \theta = \frac{e^{i\theta}- e^{-i\theta}}{ e^{i\theta}+ e^{-i\theta}}$$But your approach is far more elegant. Very nicely doen indeed! (Yes)
The formula that I used for $\tan(m\theta)$ comes straight from de Moivre's theorem, of course, so the complex numbers were definitely there in the background.
 
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