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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

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

Grade:11

1 Answers

A
11 Points
5 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.

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