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# if cos(x-y) +cos(y-z) +cos(x-z) = -3/2 prove that sinx +siny +sinz = cosx +cosy +cosz = 0

A
11 Points
2 years ago
∵ cos (x-y) + cos (y-z) + cos (z-x) = -3/2

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

∴ (sin² x + cos² x) + (sin² y + cos² y) + (sin² z + cos² z)
. . .+ 2 cos x cos y + 2 sin x sin y
. . .+ 2 cos y cos z + 2 sin y sin z
. . .+ 2 cos z cos x + 2 sin z sin x = 0

∴ ( sin² x + sin² y + sin² z + 2 sin x sin y + 2 sin y sin z + 2 sin z sin x )
. . .+( cos² x + cos² y + cos² z + 2 cos x cos y + 2 cos y cos z + 2 cos z cos x ) = 0

∴ (sin x + sin y + sin z )² + ( cos x + cos y + cos z )² = 0 ................... (1)

... But a² + b² = 0 iff a = 0 and b = 0.

∴ sin x + sin y + sin z = 0 AND cos x + cos y + cos z = 0.

∴ cos x + cos y + cos z = 0 = sin x + sin y + sin z ............... Q.