Math, asked by Anonymous, 7 months ago

Question 19 as soon as possible.

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Answered by Anonymous
16

Answer:

\sf cos(x-y)+cos(y-z) +cos(x-z)cos(x−y)+cos(y−z)+cos(x−z)

2 (cos x cos y + sin x sin y + cos y cos z + sin y sin z + cos x cos z + sin x sin z) = -3

3 + 2 (cos x cos y + cos y cos z + cos x cos z) + 2(sin x sin y + sin y sin z + sin x sin z) = 0

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

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

(sin x + sin y + sin z)² + (cos x + cos y + cos z)² = 0

(sin x + sin y + sin z)² = (cos x + cos y + cos z)² = 0

sin x + sin y + sin z = cos x + cos y + cos z = 0

HENCE \sum \rm cos(x) = cos(x) + cos(y) + cos(z)∑cos(x)=cos(x)+cos(y)+cos(z) = 0

Answered by Anonymous
14

option D _____________

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