Trigo. product (1 Viewer)

asianese · Feb 5, 2013

Any hint?

Lieutenant_21 · Feb 5, 2013

$I got the result for this (using z^7-1=0):$ $cos\frac{\pi }{7}.cos\frac{2\pi }{7}.cos\frac{3\pi }{7}$
is there a way to get the result for your question from there? I don't think so

deswa1 · Feb 5, 2013

Hahaha I've done this question before (its either past HSC or in Terry Lee) but they led you through it massively. William's point of using the roots of z^7-1=0 is right I think. I'll try find the exact question).

asianese · Feb 5, 2013

Hmm tired z^7-1=0 also.. Terry Lee has Page 78: cos(pi/7)cos(2pi/7)cos(3pi/7) = 1/8.

RealiseNothing · Feb 5, 2013

An easy way to get the cosine result without using $z^7-1$ is:

$sin(\frac{\pi}{7})=sin(\frac{6\pi}{7})$

$sin(\frac{3\pi}{7})=sin(\frac{4\pi}{7})$

So that means that:

$sin(\frac{\pi}{7})sin(\frac{2\pi}{7})sin(\frac{3\pi}{7})=sin(\frac{2\pi}{7})sin(\frac{4\pi}{7})sin(\frac{6\pi}{7}$

$\frac{sin(\frac{\pi}{7})sin(\frac{2\pi}{7})sin(\frac{3\pi}{7})}{sin(\frac{2\pi}{7})sin(\frac{4\pi}{7})sin(\frac{6\pi}{7})}=1$

Using the identity on each respective sine:

$\frac{sin2\theta}{sin\theta}=2cos\theta$

$\frac{sin(\frac{2\pi}{7})sin(\frac{4\pi}{7})sin(\frac{6\pi}{7})}{sin(\frac{\pi}{7})sin(\frac{2\pi}{7})sin(\frac{3\pi}{7})}{}=8cos(\frac{\pi}{7})cos(\frac{2\pi}{7})cos(\frac{3\pi}{7})=1$

$cos(\frac{\pi}{7})cos(\frac{2\pi}{7})cos(\frac{3\pi}{7})=\frac{1}{8}$

RealiseNothing · Feb 5, 2013

lol all my latex is right, it doesn't seem to want to show up properly though in some spots.

Lieutenant_21 · Feb 18, 2013

$Can we do it by evaluating the following product:$

$\prod_{k=0}^6 \sin(k\pi/7)$

We can do this by expressing the sine as complex exponentials.

$\frac{z^7-1}{z-1}=\prod_{k=1}^6 (1- \zeta^k)$

$with$ $\zeta = \cos(\pi/7) + i\sin(\pi/7)$ . $Let z go to 1.$

Trigo. product (1 Viewer)

juantheron

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asianese

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deswa1

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asianese

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RealiseNothing

what is that?It is Cowpea

RealiseNothing

what is that?It is Cowpea

Lieutenant_21

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