Evaluate $ \cos(x) \cos(2x) \cos(3x) ... \cos(999x) $ where $x = \frac{2 \pi}{ 1999 }$.

Asked Nov 18 '19 at 07:47

Active Nov 18 '19 at 07:47

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Evaluate

$$ \cos(x) \cos(2x) \cos(3x) ... \cos(999x) $$

where $x = \frac{2 \pi}{ 1999 }$.

Attempt:

Let $A = \cos(x) \cos(2x) \cos(3x) ... \cos(999x)$, $B = \sin(x) \sin(2x) \sin(3x) ... \sin(999x)$. Then

$$ AB = \frac{\sin(2x) \sin(4x) ... \sin(1998x)}{2^{999}} $$

then

$$ A = \frac{\sin(2x) \sin(4x) ... \sin(1998x)}{2^{999}(\sin(x) \sin(2x) ... \sin(999x))} $$ $$ = \frac{\sin(1000x) \sin(1002x) ... \sin(1998x)}{2^{999}(\sin(x) \sin(3x)... \sin(997x) \sin(999x))} $$

What to do after this? If I can find $1999x = 2\pi$ then it would be nice.

asked Nov 18 '19 at 07:47

Redsbefall

4,845

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$sin (2\pi-x) = - sin (x) $ – Someone Nov 18 '19 at 07:52
1

https://math.stackexchange.com/questions/1284744/find-the-value-of-cos-x-cos-2x-cos-999x-given-that-x-frac-2-pi1999 – lab bhattacharjee Nov 18 '19 at 07:57
https://math.stackexchange.com/questions/1351337/product-of-cosines-prod-r-17-cos-left-fracr-pi15-right – Claude Leibovici Nov 18 '19 at 07:58

Evaluate $ \cos(x) \cos(2x) \cos(3x) ... \cos(999x) $ where $x = \frac{2 \pi}{ 1999 }$.

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