Cos (a-b) +cos (b-c) + cos (c-a) = -3/2Prove that cos a + cos b+ cos c = sin a + sin b + sin c = 0
Kritika , 9 Years ago
Grade 10
Trigonometry> Cos (a-b) +cos (b-c) + cos (c-a) = -3/2 P...
1 Answersaditi arora
Last Activity: 9 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)
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