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Evaluate Trigonometric Expression Challenge

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Evaluate \(\displaystyle \tan\frac{\pi}{13}\tan\frac{2\pi}{13}\tan\frac{3 \pi}{13}\tan\frac{4\pi}{13}\tan\frac{5\pi}{13} \tan \frac{6\pi}{13}\).
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
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Re: Evaluate the expression of trigonometric

Evaluate \(\displaystyle \tan\frac{\pi}{13}\tan\frac{2\pi}{13}\tan\frac{3 \pi}{13}\tan\frac{4\pi}{13}\tan\frac{5\pi}{13} \tan \frac{6\pi}{13}\).
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 $$\tan\bigl(\tfrac{\pi}{13}\bigr) \tan\bigl(\tfrac{2\pi}{13}\bigr) \tan\bigl(\tfrac{3\pi}{13}\bigr) \tan\bigl(\tfrac{4\pi}{13}\bigr) \tan\bigl(\tfrac{5\pi}{13}\bigr) \tan\bigl(\tfrac{6\pi}{13}\bigr) = \sqrt{13}.$$
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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}$ ?
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: Evaluate the expression of trigonometric

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 $\tan(\pi/3) = \sqrt3$.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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}$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 this thread.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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}}$