Finite Tangent product / quotient

[DreamWeaver]

Just for fun, eh...? (Heidy)For \(\displaystyle z \in \mathbb{R}\), and \(\displaystyle m \in 2\mathbb{N}+1\), show that:\(\displaystyle \frac{\tan mz}{\tan z}=\prod_{j=1}^{ \lfloor m/2 \rfloor } \tan\left(\frac{j\pi}{m}+z\right) \tan\left(\frac{j\pi}{m}-z\right) \)

 

[Opalg]

[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]

 

[DreamWeaver]

@ Opalg...

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

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

 

[Opalg]

@ Opalg...

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

\(\displaystyle \tan \theta = \frac{e^{i\theta}- e^{-i\theta}}{ e^{i\theta}+ e^{-i\theta}}\)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.

 

[CellCoree]

Interesting question, Heidy! This equation deals with the finite tangent product and quotient, which is a mathematical concept used to calculate the product or quotient of multiple tangent functions. In this case, we are looking at the specific scenario where z is a real number and m is an odd integer.

To prove this equation, we can start by rewriting the left side of the equation as a product of tangent functions:

\frac{\tan mz}{\tan z} = \frac{\sin mz}{\cos mz} \cdot \frac{\cos z}{\sin z} = \frac{\sin mz \cos z}{\cos mz \sin z}

Next, we can use the double angle formula for tangent to rewrite the numerator and denominator:

\sin mz \cos z = \frac{1}{2}(\sin (m+1)z + \sin (m-1)z) \cdot \frac{1}{2}(\sin z + \sin z) = \frac{1}{4}\sin (m+1)z \sin z + \frac{1}{4}\sin (m-1)z \sin z

\cos mz \sin z = \frac{1}{2}(\cos (m+1)z - \cos (m-1)z) \cdot \frac{1}{2}(\sin z + \sin z) = \frac{1}{4}\cos (m+1)z \sin z - \frac{1}{4}\cos (m-1)z \sin z

Plugging these back into the original equation, we get:

\frac{\sin mz \cos z}{\cos mz \sin z} = \frac{\frac{1}{4}\sin (m+1)z \sin z + \frac{1}{4}\sin (m-1)z \sin z}{\frac{1}{4}\cos (m+1)z \sin z - \frac{1}{4}\cos (m-1)z \sin z}

Now, we can use the sum and difference identities for sine and cosine to simplify the numerator and denominator:

\frac{\sin mz \cos z}{\cos mz \sin z} = \frac{\frac{1}{4}\sin mz \sin z}{\frac{1}{4}\cos mz \sin z} = \frac{\tan mz}{\