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Trignometry

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12 Oct 2009 11:48:09 IST

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

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

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aal izz well

Blazing goIITian

Joined: 6 Jun 2009

Posts: 409

12 Oct 2009 18:53:35 IST

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is the answer 64

Millind Gupta

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Joined: 7 Oct 2009

Posts: 227

13 Oct 2009 20:48:56 IST

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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)

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

akki ~~ unlucky forever ~~

Blazing goIITian

Joined: 11 May 2008

Posts: 491

16 Oct 2009 18:50:02 IST

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here's another method :

multiply numerator & denominator by (2cosx+1)

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

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

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