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Pls solve.
Asked by pb_ckt | 11 Jun, 2019, 03:34: PM
sin x + sin y + sin z = 0 and cos x + cos y + cos z = 0
Hence,
(sin x + sin y + sin z)2 + (cos x + cos y + cos z)2 = 0
sin2 x + sin2 y + sin2 z + 2(sin x + sin y + sin z) + cos2 x + cos2 y + cos2 z + 2(cos x + cos y + cos z) = 0
sin2 x + cos2 x + sin2 y + cos2 y + sin2 z + cos2 z + 2(cos x cos y + sin x sin y + cos y cos z + sin y sin z + cos z cos x + sin z sin x) = 0
1 + 1 + 1 + 2[cos (x - y) + cos (y - z) + cos (z - x)] = 0
3 + 2[cos (x - y) + cos (y - z) + cos (z - x)] = 0

cos (x - y) + cos (y - z) + cos (z - x) = -3/2
Answered by Sneha shidid | 12 Jun, 2019, 10:42: AM

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