Prove matrix det(cos x cos z 1, cos y 1 cos z, 1 cos y cos x) = 0 if x+y+z= 2 pi

Question

Prove matrix det(cos x cos z 1, cos y 1 cos z, 1 cos y cos x) = 0 if x+y+z= 2 pi

Answer 1

Hi Vin C.,

Its probably too late but anyway.

The det. = cos(x)*[ 1*cos(x) - cos(y)*cos(z)] - cos(z)*[cos(x)*cos(y) - cos(z)*1] + 1*[cos(y)*cos(y) - 1*1]

= cos²(x) - cos(x)*cos(y)*cos(z) - cos(x)*cos(y)*cos(z) + cos²(z) + cos²(y) - 1

= cos²(x) + cos²(y) + cos²(z) - 2*cos(x)*cos(y)*cos(z) - 1

Using half-angle identity: cos²(u) = 1/2 + cos(2u)/2

cos²(x) + cos²(y) + cos²(z) =

[1/2 + cos(2x)/2] + [1/2 + cos(2y)/2] + [1/2 + cos(2z)/2] =

3/2 +[cos(2x) +cos(2y) + cos(2z)]/2

Using product-to-sum formula: cos(u)*cos(v) = [cos(u+v) +cos(u-v)]/2

2*cos(x)*cos(y)*cos(z) =

2*[cos(x+y) + cos(x-y)]*cos(z)/2 =, ....(2's cancel)

cos(x+y)*cos(z) + cos(x-y)*cos(z) =

[cos(x+y+z) + cos(x+y-z) + cos(x-y+z) + cos(x-y-z)]/2 = ,...(x+y+z=2π; x+y=2π-z; x+z=2π-y; -y-z=x-2π)

[cos(2π) + cos(2π-2z) + cos(2π-2y) + cos(2x-2π)]/2 =, ...(add/sub formulas)

[1 + cos(2x) + cos(2y) +cos(2z)]/2

Combine:

3/2 +[cos(2x) +cos(2y) + cos(2z)]/2 - [1 + cos(2x) + cos(2y) +cos(2z)]/2 -1 = 0

I hope this helps, Joe.