MQ56 --- Trigonometry

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圖片參考：http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-1820.jpg

Consider:
cos(0) + cos(2x) + cos(4x) + ... + cos(10x) ... (i)
By cosx = [e^ix + e^(-ix)]/2
Let π/11 = x ... (ii)
2 x (i)
= 1 + e^2ix + e^4ix + ... + e^10ix) + (1 + e^(-2ix) + e^(-4ix) + ... + e^(-10ix)
which are geometric sequences
= 1{1 - [e^(2ix)]^6}/[1 - e^(2ix)] + 1{1 - [e^-(2ix)]^6}/[1 - e^(-2ix)]
= [1 - e^(12ix)]/[1 - e^(2ix)] + [1 - e^(-12ix)]/[1 - e^(-2ix)]
= {[1 - e^(-2ix) - e^(12ix) + e^(10ix)] + [1 - e^(2ix) - e^(-12ix) + e^(10ix)]} / [2 - e^(2ix) - e^(-2ix)]
= [2 - 2cos2x - 2(cos12x - cos10x)] / (2 - 2cos2x)
= 1 - (cos12x - cos10x) / (1 - cos2x)
= 1 + 2sin(11x)sinx/(1 - cos2x)
= 1 + 2sin(π)sinx/(1 - cos2x) ... by (ii)
= 1 + 0
= 1

i.e.
cos(0) + cos(2x) + ... + cos(10x) = (i)/2 = 1/2
2 + 2cos(2x) + 2cos(4x) + ... + 2cos(10x) = 1
2cos(2x) + 2cos(4x) + ... + 2cos(10x) + 1 = 0
2cos(2x) + 4cos(4x) + 2cos(6x) + 2cos(10x) + 1 = 2cos(4x) -2cos(8x)
2[cos(10x) + cos(2x)] + 4cos(4x) + 2cos(6x) + 1 = -2[cos(8x) - cos(4x)]
[sum to product]
4cos(6x)cos(4x) + 4cos(4x) + 2cos(6x) + 1 = 4sin(6x)sin(2x)

3[cos(6x)cos(4x) + cos(4x) + cos(6x) + 1] + [cos(6x)cos(4x) + cos(4x) - cos(6x) - 1] - 1 = 4sin(6x)sin(2x)

3[cos(6x) + 1][cos(4x) + 1] + [cos(6x) + 1][cos(4x) - 1] - 1 = 4sin(6x)sin(2x)
[double angle formulae]
3[2cos^2(3x)][2cos^2(2x)] + [2cos^2(3x)][-2sin^2(2x)] - 1 = 4[2sin(3x)cos(3x)]sin(2x)

12cos^2(3x)cos^2(2x) - 4cos^2(3x)sin^2(2x) - 1 = 8sin(3x)cos(3x)sin(2x)

12cos^2(3x)cos^2(2x) + 12cos^2(3x)sin^2(2x) - 16cos^2(3x)sin^2(2x) - 1
= 8sin(3x)cos(3x)sin(2x)

12cos^2(3x)(1)
= 1 + 16cos^2(3x)sin^2(2x) + 8sin(3x)cos(3x)sin(2x)

11cos^2(3x)
= [1 - cos^2(3x)] + 16cos^2(3x)sin^2(2x) + 8sin(3x)cos(3x)sin(2x)
= sin^2(3x) + 8sin(3x)cos(3x)sin(2x) + 16cos^2(3x)sin^2(2x)
= sin^2(3x) + 2sin(3x)[4cos(3x)sin(2x) + [4cos(3x)sin(2x)]^2
= [sin(3x) + 4cos(3x)sin(2x)]^2

11
= {[sin(3x) + 4cos(3x)sin(2x)]^2} / cos^2(3x)
= [sin(3x)/cos(3x) + 4cos(3x)sin(2x)/cos(3x)]^2
= [tan(3x) + 4sin(2x)]^2

i.e.
[tan(3π/11) + 4sin(2π/11)]^2 = 11 , By (ii)
tan(3π/11) + 4sin(2π/11) = 11^(1/2) since (3π/11 < π/2) and (2π/11 < π/2)