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Cos (a-b) +cos (b-c) + cos (c-a) = -3/2 Prove that cos a + cos b+ cos c = sin a + sin b + sin c = 0

Cos (a-b) +cos (b-c) + cos (c-a) = -3/2
Prove that cos a + cos b+ cos c = sin a + sin b + sin c = 0 

Grade:10

1 Answers

aditi arora
29 Points
7 years ago
Cos (a-b) +cos (b-c) + cos (c-a) = -3/2
multiply the equation by 2 then add 3 so that rhs becomes 0 (by doing this it becomes easy for us to solve)
now expand the equation by using cos (a-b)=cos a cos b + sin a sin b
when you do so it becomes
(cos a + cos b + cos +c )2 + (sin a + sin b + sin c)2 = 0
this is possible only when (cos a + cos b +cos c) = 0 = (sin a+ sin b+ sinc)
HENCE PROOVED

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