Evaluate $ \cos a \cos 2 a \cos 3 a \cdots \cos 999 a $ where $a=\frac{2 \pi}{1999}$

Question

Evaluate $ \cos a \cos 2 a \cos 3 a \cdots \cos 999 a $ where $a=\frac{2 \pi}{1999}$

I know this question has already been answered many times but my doubt is different

Solution: Let $P$ denote the desired product, and let

$ Q=\sin a \sin 2 a \sin 3 a \cdots \sin 999 a $

Then $ \begin{aligned} 2^{999} P Q=&(2 \sin a \cos a)(2 \sin 2 a \cos 2 a) \cdots(2 \sin 999 a \cos 999 a) \\ =& \sin 2 a \sin 4 a \cdots \sin 1998 a \\ =&(\sin 2 a \sin 4 a \cdots \sin 998 a)[-\sin (2 \pi-1000 a)] \\ & \cdot[-\sin (2 \pi-1002 a)] \cdots[-\sin (2 \pi-1998 a)] \\ =& (\sin 2 a \sin 4 a \cdots \sin 998 a) \sin 999 a \sin 997 a \cdots \sin a=Q \end{aligned} $

how they got this last step from previous one ???

Answer 1

To reach penultimate step, use $\sin(2\pi-\theta)=-\sin\theta$, and to go from penultimate to ultimate step, use $$2\pi-1000a=2\pi-1000\cdot\frac{2\pi}{1999}=2\pi\left(1-\frac{1000}{1999}\right)=2\pi\frac{999}{1999}=999a$$ Also, there comes out $(-1)^{999-500+1}$ if one counts only even numbers from $1000$ to $1998$.