y=(2cosx-1)(2cos2x-1)(2cos4x-1)------(2cos32x-1) find y at x=2pi/13

akki ~~ unlucky forever ~~

here's another method :

multiply numerator & denominator by (2cosx+1)

$y=\frac{(4cos^2x-1)(2cos2x-1)(2cos4x-1)...(2cos32x-1)}{(2cosx+1)}\\ \\ y=\frac{(2(2cos^2x-1)+1)(2cos2x-1)(2cos4x-1)...(2cos32x-1)}{(2cosx+1)}\\ \\ y=\frac{(2cos2x+1)(2cos2x-1)(2cos4x-1)...(2cos32x-1)}{(2cosx+1)}\\ \\ in\ same\ way\ the\ numerator\ goes\ on\ simplifying\ to\ \\ y=\frac{(4cos^232x-1)}{(2cosx+1)}\\ \\ y=\frac{(2cos64x+1)}{(2cosx+1)}$

$at\ x=\frac{2\pi}{13}\\ cos64x=cos(10\pi-(\frac{2\pi}{13}))=cos\frac{2\pi}{13}\\ y = \frac{(2cos\frac{2\pi}{13}+1)}{(2cos\frac{2\pi}{13}+1)}=1$