## Monday, October 31, 2011

### Product of Multiple Trignometry Function

PMTF.ID.1:   Product of multiple sine
$\prod _{k=1}^{n-1}\sin (\frac{k\pi }{n})= \frac{n}{2^{n-1}}$

Proof
$P= \prod _{k=1}^{n-1}\sin (\frac{k\pi }{n})= (2i)^{1-n}\prod _{k=1}^{n-1} \left ( e^{\frac{i\pi k}{n}}- e^{\frac{-i\pi k}{n}} \right )$
(1)
$P= (2i)^{1-n} e^{\frac{-\pi i}{n} }e^{\frac{-2\pi i }{n} }\cdot \cdot \cdot e^{\frac{-k\pi i }{n} }\cdot \cdot \cdot e^{\frac{-(n-1)\pi i }{n} } \prod _{k=1}^{n-1} (e^{\frac{2\pi i k }{n} }-1)$
(2)
$P= (2i)^{1-n} e^{-\frac{\pi i \sum _{k=1}^{n-1}k }{n} } \prod _{k=1}^{n-1} (e^{\frac{2\pi i k }{n} }-1)$
(3)

$\sum _{k=1}^{n-1}k=\frac{n(n-1)}{2}$
(4)
from (3) and (4) ,

$P= (2i)^{1-n} e^{-\frac{\pi i n(n-1) }{2n} } \prod _{k=1}^{n-1} (e^{\frac{2\pi i k }{n} }-1)$
(5)
$i=e^{\frac{\pi i }{2}}$

(6)

from (6) and (5)
$P= (-2)^{1-n}e^{\frac{\pi i (1-n) }{2}} e^{-\frac{\pi i n(n-1) }{2n} } \prod _{k=1}^{n-1} (e^{\frac{2\pi i k }{n} }-1)$
(7)
$P= (-2)^{1-n} \prod _{k=1}^{n-1} (e^{\frac{2\pi i k }{n} }-1)$

(8)
$P= (2)^{1-n} \prod _{k=1}^{n-1} (1-e^{\frac{2\pi i k }{n} })$
(9)
$\varepsilon =e^{\frac{2\pi i }{n}} \, ,\, \varepsilon^{k} =e^{\frac{2\pi ik }{n}} \,$
(10)
$\frac{x^{n}-1}{x-1} = \sum _{k=0}^{n-1}x^{k}$
(11)
$x^{n}-1 = (x-1)\sum _{k=0}^{n-1}x^{k}$
(12)
the equation (12) is represented as product of zeros , shown as following
$x^{n}-1 = \prod _{k=0}^{n-1}(x-\varepsilon ^{k})$
(13)

$x^{n}-1 = (x-1)\prod _{k=1}^{n-1}(x-\varepsilon ^{k})$

(14)
from (12) and (14)
$(x-1)\sum _{k=0}^{n-1}x^{k} = (x-1)\prod _{k=1}^{n-1}(x-\varepsilon ^{k})$
(15)
$\sum _{k=0}^{n-1}x^{k} = \prod _{k=1}^{n-1}(x-\varepsilon ^{k})$
(16)
if x=1 , then

$n = \prod _{k=1}^{n-1}(1-\varepsilon ^{k})$
(17)
from (17) and (9)

$P= 2^{1-n}n$
(18)
$P= \frac{n}{2^{n-1}}$
(19)

PMTF.ID.1:   Product of multiple cosine

$\prod _{k=1}^{n-1}\cos (2^{k}\alpha )= \frac{\sin (2^{n}\alpha )}{2^{n}\sin (\alpha )}$
Proof

$\sin (2\alpha )=2\sin (\alpha )\cos (\alpha )$
(1)

$\cos (\alpha )= \frac{\sin (2\alpha )}{2\sin (\alpha )}$
(2)
$\cos (2\alpha )= \frac{\sin (4\alpha )}{2\sin (2\alpha )}$
(3)
$\cos (4\alpha )= \frac{\sin (8\alpha )}{2\sin (4\alpha )}$
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$\cos (2^{n-1}\alpha )= \frac{\sin (2^{n}\alpha )}{2\sin (2^{n-1}\alpha )}$
(4)
$\cos (\alpha )\cos (2\alpha )\cos (4\alpha )....\cos (2^{n-1}\alpha )= \frac{\sin (2^{n}\alpha )}{2^{n}\sin (\alpha )}$
(5)
$\prod _{k=1}^{n-1}\cos (2^{k}\alpha )= \frac{\sin (2^{n}\alpha )}{2^{n}\sin (\alpha )}$
(6)