shmm wrote: Find the product $ \cos a \cdot \cos 2a\cdot \cos 3a \cdots \cos 1006a$ where $a=\frac{2\pi}{2013}$. Let $T_{2013}(x)$ be the $2013^e$ Chebyshev polynomial. Coefficient of $x^{2013}$ is $2^{2012}$ and constant coefficient is $0$ The $2013$ roots of equation $P(x)=1$ are $\{\cos ka\}_{k=0}^{2012}$ and so $\prod_{k=0}^{2012}\cos ka=2^{-2012}$ And since $\prod_{k=0}^{2012}\cos ka=\left(\prod_{k=1}^{1006}\cos ka\right)^2$, we get $\prod_{k=1}^{1006}\cos ka=\pm 2^{-1006}$ It remains to see that this product contains exactly $503$ positive factors and exactly $503$ negative factors, and we get the result : $\boxed{\prod_{k=1}^{1006}\cos ka=-2^{-1006}}$

$P= \cos a\cdot\cos 2a\cdot\cos 3a\cdots\cos 1006a $ then $P^2= \cos a\cdot\cos 2a\cdot\cos 3a\cdots\cos 2012a $ $P^2=2^{-2012}\prod(e^{\frac{i2\pi k}{2013}}+e^{\frac{-i2\pi k}{2013}})$ from k=1 to k=2012 Now, $\prod(e^{\frac{i2\pi k}{2013}})$=product of the complex 2013th root of unity=-1 Also,$ \prod(e^{\frac{i4\pi k}{2013}}+1)=\prod(e^{\frac{i2\pi k}{2013}+1})$ for k=1 to k=2012, since ${2*1, 2*2, 2*3, 2*4, 2*5, .............2*2012}$ form a complete residue system $modulo 2013$ Also, $\prod(e^{\frac{i2\pi k}{2013}}-x)=x^{2012}+x^{2011}+.........+x^2+x+1$ putting $x=-1$ we get the value of $P^2$