MHB Evaluate Trigonometric Expression Challenge

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The discussion centers on evaluating the product of tangent functions: $$\tan\frac{\pi}{13}\tan\frac{2\pi}{13}\tan\frac{3 \pi}{13}\tan\frac{4\pi}{13}\tan\frac{5\pi}{13} \tan \frac{6\pi}{13}$$, which is determined to equal $\sqrt{13}$. This result stems from analyzing the roots of the equation $\tan(13x) = 0$ and applying polynomial identities. Participants also explore the possibility of generalizing this result to the product of tangents for any integer $n$, concluding that $$\prod_{k=1}^{n} \tan \left( \frac{k \pi}{2n+1} \right) = \sqrt{2n+1}$$ holds true. Additionally, they discuss evaluating the products of sine and cosine functions in closed form, establishing relationships between these trigonometric functions. The conversation highlights significant connections in trigonometric identities and their implications in broader mathematical contexts.
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Evaluate $$\tan\frac{\pi}{13}\tan\frac{2\pi}{13}\tan\frac{3 \pi}{13}\tan\frac{4\pi}{13}\tan\frac{5\pi}{13} \tan \frac{6\pi}{13}$$.
 
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Re: Evaluate the expression of trigonometric

anemone said:
Evaluate $$\tan\frac{\pi}{13}\tan\frac{2\pi}{13}\tan\frac{3 \pi}{13}\tan\frac{4\pi}{13}\tan\frac{5\pi}{13} \tan \frac{6\pi}{13}$$.
[sp]The numbers $\frac{k\pi}{13}\ (0\leqslant k\leqslant 12)$ are the roots of the equation $\tan(13x) = 0.$ But $$\tan(13x) = \frac{\sin(13x)}{\cos(13x)} = \frac{{13\choose1}t - {13\choose3}t^3 + {13\choose5}t^5 - \ldots + t^{13}}{ 1 - {13\choose2}t^2 + {13\choose4}t^4 - \ldots + {13\choose12}t^{12}},$$ where $t = \tan x$ (as you can see by applying De Moivre's theorem to $(\cos x + i\sin x)^{13}$). The equation $\tan(13x) = 0$ will hold when the numerator of that fraction is $0$, in other words when $${13\choose1}t - {13\choose3}t^3 + {13\choose5}t^5 - \ldots + t^{13} = 0. \qquad(*)$$ Therefore the roots of (*) are $\tan\bigl(\frac{k\pi}{13}\bigr)\ (0\leqslant k\leqslant 12)$. One of the roots is $\tan 0 = 0$. Dividing by $t$ to get rid of that root, we are left with the equation $${13\choose1} - {13\choose3}t^2 + {13\choose5}t^4 - \ldots + t^{12} = 0, \qquad(**)$$ whose roots are $\tan\bigl(\frac{k\pi}{13}\bigr)\ (1\leqslant k\leqslant 12)$. The product of the roots of (**) is the constant term, $13$. But $\tan\bigl(\frac{(13-k)\pi}{13}\bigr) = -\tan\bigl(\frac{k\pi}{13}\bigr)$, so the product of the 12 roots is the square of the product of the first six roots. Therefore $$\tan\bigl(\tfrac{\pi}{13}\bigr) \tan\bigl(\tfrac{2\pi}{13}\bigr) \tan\bigl(\tfrac{3\pi}{13}\bigr) \tan\bigl(\tfrac{4\pi}{13}\bigr) \tan\bigl(\tfrac{5\pi}{13}\bigr) \tan\bigl(\tfrac{6\pi}{13}\bigr) = \sqrt{13}.$$[/sp]
 
Re: Evaluate the expression of trigonometric

@Opalg

Is there any reason we couldn't generalize and conclude $ \displaystyle \prod_{k=1}^{n} \tan \left( \frac{k \pi}{2n+1} \right) = \sqrt{2n+1}$ ?
 
Last edited:
Re: Evaluate the expression of trigonometric

Random Variable said:
Is there any reason we couldn't generalize and conclude $ \displaystyle \prod_{k=1}^{n} \tan \left( \frac{k \pi}{2n+1} \right) = \sqrt{2n+1}$ ?
That is correct. (Yes) It even works when $n=1$, to give $\tan(\pi/3) = \sqrt3$.
 
Can we also evaluate $ \displaystyle \prod_{k=1}^{n} \sin \left( \frac{k \pi}{2n+1} \right)$ and $ \displaystyle \prod_{k=1}^{n} \cos \left( \frac{k \pi}{2n+1} \right) $ in closed form?

We know that $ \displaystyle \frac{\prod_{k=1}^{n} \sin \left( \frac{k \pi}{2n+1} \right)}{\prod_{k=1}^{n} \cos \left( \frac{k \pi}{2n+1} \right)} = \sqrt{2n+1}$.
 
Random Variable said:
Can we also evaluate $ \displaystyle \prod_{k=1}^{n} \sin \left( \frac{k \pi}{2n+1} \right)$ and $ \displaystyle \prod_{k=1}^{n} \cos \left( \frac{k \pi}{2n+1} \right) $ in closed form?

We know that $ \displaystyle \frac{\prod_{k=1}^{n} \sin \left( \frac{k \pi}{2n+1} \right)}{\prod_{k=1}^{n} \cos \left( \frac{k \pi}{2n+1} \right)} = \sqrt{2n+1}$.
See http://www.mathhelpboards.com/f12/simplify-cos-cos-2a-cos-3a-cos-999a-if-%3D-2pi-1999-a-253/#post1517.
 
I came up with something for the direct evaluation of $ \displaystyle \prod_{k=1}^{n} \sin \left( \frac{k \pi}{2n+1} \right)$.

I'm first going to show that $ \displaystyle \prod_{k=1}^{2n} \Bigg( 1-\exp \left(\frac{2 \pi i k}{2n+1} \right) \Bigg)= (2n+1)$.

$ \displaystyle z^{2n+1}-1 = \prod_{k=0}^{2n} \Bigg( z- \exp \left( \frac{2 \pi i k}{2n+1} \right) \Bigg) = (z-1) \prod_{k=1}^{2n} \Bigg( z- \exp \left( \frac{2 \pi i k}{2n+1}\right) \Bigg)$

$ \displaystyle \implies \prod_{k=1}^{2n} \Bigg( z-\exp \left(\frac{2 \pi i k}{2n+1} \right) \Bigg) = \frac{z^{2n+1}-1}{z-1}$

$ \displaystyle \implies \prod_{k=1}^{2n} \Bigg( 1-\exp \left(\frac{2 \pi i k}{2n+1} \right) \Bigg) = \lim_{z \to 1} \frac{z^{2n+1}-1}{z-1} = \lim_{z \to 1} \frac{(2n+1) z^{2n}}{1} = 2n+1$ Using the fact $ \displaystyle \exp \left(\frac{\pi i k }{2n+1} \right) \sin \left(\frac{k \pi}{2n+1} \right) = \frac{i}{2} \Bigg( 1- \exp \left(\frac{2\pi i k}{2n+1} \right) \Bigg)$

$ \displaystyle \prod_{k=1}^{2n} \exp \left(\frac{\pi i k }{2n+1} \right) \sin \left(\frac{k \pi}{2n+1} \right) = \prod_{k=1}^{2n} \exp \left(\frac{\pi i k }{2n+1} \right) \prod_{k=1}^{2n} \sin \left(\frac{ k \pi}{2n+1} \right) $

$ \displaystyle = \Big[ \exp \left( \frac{\pi i}{2n+1} \right) \Big]^{n(2n+1)}\prod_{k=1}^{2n} \sin \left(\frac{k \pi}{2n+1} \right) = \displaystyle (-1)^{n} \prod_{k=1}^{2n} \sin \left(\frac{k \pi}{2n+1} \right)$

$ \displaystyle = \prod_{k=1}^{2n} \frac{i}{2} \Bigg( 1- \exp \left(\frac{2\pi i k}{2n+1} \right) \Bigg) = \frac{(-1)^{n}}{2^{2n}} (2n+1)$

$\displaystyle \implies \prod_{k=1}^{2n} \sin \left(\frac{k \pi}{2n+1} \right)= \frac{2n+1}{2^{2n}}$And $ \displaystyle \prod_{k=1}^{n} \sin \left(\frac{k \pi}{2n+1} \right) = \sqrt{ \prod_{k=1}^{2n} \sin \left(\frac{k \pi}{2n+1} \right)} = \frac{\sqrt{2n+1}}{2^{n}}$
 

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