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標題:

PROVE IDENTITY

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Prove following identity... 1+cos2x+cos2y+cos2z=-4cosxcosycosz 更新: If x + y + z = π is not given, does the result hold?

最佳解答:

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.

其他解答:

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