is the answer 64

Hi dear

at last i got it solved after struggling for 3 days

here goes the solution

f(x) = (2cosx-1)(2cos2x-1)(2cos4x-1)(2cos8x-1)(2cos16x-1)(2cos32x-1)

f(2pi/13) = (2cosx-1)(2cos2x-1)(2cos4x-1)(2cos5x-1)(2cos3x-1)(2cos6x-1)   where x = 2pi/13

as  16x = 32pi/13 = 2pi + 6pi/13          which gives    cos(16x)  = cos(6pi/13)  = cos(3x)

       8x  = 16pi/13 =  2pi - 10pi/13                                   cos(8x)  = cos(10pi/13) = cos(5x)                                             

       32x = 64pi/13 = 4pi + 12pi/13                                  cos(32x) = cos(12pi/13) = cos(6x)

f(x) = (2cosx-1)(2cos2x-1)(2cos3x-1)(2cos4x-1)(2cos5x-1)(2cos6x-1)    where x = 2pi/13

let us find an equation whose roots are cos(2rpi/13) where r = 1,2,3,4,5,6

2cosx = z + 1/z    ( = v, let say)                     where z = cosx + i sinx

2cos2x = z² + 1/z²       =  v² - 2

2cos3x = z³ + 1/z³      = v³ - 3v

2cos4x = z⁴ + 1/z⁴      = v⁴ - 4v² + 2

2cos5x = z⁵ + 1/z⁵      = v⁵ - 5v³ + 5v

2cos6x = z⁶ + 1/z⁶     = v⁶ - 6v⁴ + 9v² - 2

z⁶ + 1/z⁶ + z⁵ + 1/z⁵ + z⁴ + 1/z⁴ + z³ + 1/z³ + z² + 1/z² + z + 1/z + 1 = v⁶ + v⁵ - 5v⁴ - 4v³ + 6v² + 3v - 1

(2cosx - v)(2cos2x - v)(2cos3x - v)(2cos4x - v)(2cos5x - v)(2cos6x - v) = v⁶ + v⁵ - 5v⁴ - 4v³ + 6v² + 3v - 1

put v =1

(2cosx-1)(2cos2x-1)(2cos3x-1)(2cos4x-1)(2cos5x-1)(2cos6x-1)  = 1 + 1 - 5 - 4 + 6 + 3 - 1 = 1

Answer

I've already proved that for x = 2pi/13

(2cosx-1)(2cos2x-1)(2cos3x-1)(2cos4x-1)(2cos5x-1)(2cos6x-1) =

(2cosx-1)(2cos2x-1)(2cos4x-1)(2cos8x-1)(2cos16x-1)(2cos32x-1)

so

(2cosx-1)(2cos2x-1)(2cos4x-1)(2cos8x-1)(2cos16x-1)(2cos32x-1)  = 1           Answer

here's another method :

multiply numerator & denominator by (2cosx+1)