21i have used sin(a) = - sin(2pi- a)
i guess if u write it completely u will understand
i wrote completely when i solved it, while posting i used this notation
\hspace{-16}$Evaluate $\mathbf{\cos (a).\cos (2a).\cos(3a).........\cos(999a)=}$\\\\ Where $\mathbf{a=\frac{2\pi}{1999}}$
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6 Answers
21We are looking for
\prod_{i=1}^{999}cos(\frac{2i\pi}{1999})= \frac{\prod_{i=1}^{999}sin(\frac{4i\pi}{1999})}{2^{999}\prod_{i=1}^{999}sin(\frac{2i\pi}{1999})}
But we have
\prod_{i=1}^{999}sin(\frac{4i\pi}{1999})= \prod_{i=1}^{499}sin(\frac{4i\pi}{1999})\prod_{i=500}^{999}sin(\frac{4i\pi}{1999})= \prod_{i=1}^{499}sin(\frac{4i\pi}{1999})(-1)^{500}\prod_{i=500}^{999}sin(\frac{(3998-4i)\pi}{1999})
=\prod_{i=1}^{999}sin(\frac{2i\pi}{1999})
Hence
\prod_{i=1}^{999}cos(\frac{2i\pi}{1999})= \frac{1}{2^{999}}
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