Best solutions to - "Given x=2π1999Find the value ofcos⁡xcos⁡2xcos⁡3x…cos⁡999x"

nomadzkia0re

Answered question

2022-03-24

Given

x = \frac{2 π}{1999}

Find the value of

\cos x \cos 2 x \cos 3 x \dots \cos 999 x

Answer & Explanation

Drake Huang

Beginner2022-03-25Added 15 answers

If $\cos (2 n + 1) x = 1, (2 n + 1) x = 2 m π$ where m is any integer
$x = \frac{2 m π}{2 n + 1}$ where $m = 0, 1, 2, \dots, 2 n$
But $\cos (2 n + 1) x = 2^{2 n} \cos^{2 n + 1} x + \dots + {(- 1)}^{n} \cos x - 1 = 0$ are $\cos \frac{2 m π}{2 n + 1}$ where $m = 0, 1, 2, \dots, 2 n$
$\Rightarrow \prod_{m = 0}^{2 n} \cos \frac{2 m π}{2 n + 1} = 1 \Rightarrow \prod_{m = 1}^{2 n} \cos \frac{2 m π}{2 n + 1} = \frac{1}{2^{2 n}}$
Again, $\cos \frac{(2 + 1 - m) 2 π}{2 n + 1} = \cos (2 π - \frac{2 m π}{2 n + 1}) = \cos \frac{2 m π}{2 n + 1}$
$\Rightarrow \prod_{m = 1}^{2 n} \cos \frac{2 m π}{2 n + 1} = {(\prod_{m = 1}^{n} \cos \frac{2 m π}{2 n + 1})}^{2}$
Hence, $\prod_{m = 1}^{n} \cos \frac{2 m π}{2 n + 1} = \pm \frac{1}{2^{n}}$
Now $\cos \frac{2 m π}{2 n + 1} < 0$ if $\frac{3 π}{2} > \frac{2 m π}{2 n + 1} > \frac{π}{2}$
But $\frac{3 (2 n + 1)}{4} > n$
The following are $n - [\frac{2 n + 1}{4}]$ negative antecedents that will determine the sign of $\prod_{m = 1}^{n} \cos \frac{2 m π}{2 n + 1}$

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