if cos(b - c) + cos(c - a) + cos(a - b) = -3/2 , then show that cos a + cos b + cos c = 0 = sin a +sin b + sin c

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Naveen · Answer

cos(A−B)+cos(B−C)+cos(C−A)=−3/2 cosA⋅cosB+sinA⋅sinB+cosB⋅cosC+sinB⋅sinC+cosC⋅cosA+sinC⋅sinA=−3/22(cosa.cosb+sina.sinb+cosb.cosc+sinb.sinc+cosc.cosa+sinc.sina)+3=0.Now we write 3 as (cos^2 a+sin^2 a)+(cos^2 b+sin^2 b)+(cos^2 c+sin^2 c) and substitute this in. We then make use of the identity(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)to obtain that(cosa+cosb+cosc)2+(sina+sinb+sinc)2=0from which the claim follows easily.Here x^2means x square and similarly in others.

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if cos(b - c) + cos(c - a) + cos(a - b) = -3/2 , then show that cos a + cos b + cos c = 0 = sin a +sin b + sin c

if cos(b - c) + cos(c - a) + cos(a - b) = -3/2 , then show that cos a + cos b + cos c = 0 = sin a +sin b + sin c

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if cos(b - c) + cos(c - a) + cos(a - b) = -3/2 , then show that cos a + cos b + cos c = 0 = sin a +sin b + sin c

if cos(b - c) + cos(c - a) + cos(a - b) = -3/2 , then show that cos a + cos b + cos c = 0 = sin a +sin b + sin c

1 Answers

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