PROVE IDENTITY

1 + cos 2x + cos 2y + cos 2z
= 2 cos2 x + 2 cos [(2y + 2z)/2] cos [(2y - 2z)/2]
= 2 cos2 x + 2 cos (y + z) cos (y - z)
= 2 cos2 x + 2 cos (π - x) cos (y - z)
= 2 cos2 x - 2 cos x cos (y - z)
= 2 cos x [cos x - cos (y - z)]
= -4 cos x sin {[x + (y - z)]/2} sin {[x - (y - z)]/2}
= -4 cos x sin [(x + y - z)/2] sin [(x - y + z)/2]
= -4 cos x sin [(π - 2z)/2] sin [(π - 2y)/2]
= -4 cos x sin (π/2 - z) sin (π/2 - y)
= -4 cos x cos y cos z
Note: The condition x + y + z = π should be given.

1+cos2x+cos2y+cos2z=-4cosxcosycosz

係咪話
1+cos^2x+cos^2y+cos^2z=-4cosxcosycosz
OR
1+cos2x+cos2y+cos2z=-4cosxcosycosz