How'd you do this one?

Geometry Level 3

Given that c o s ( A − B ) + c o s ( B − C ) + c o s ( C − A ) = − 3 / 2 cos(A-B) + cos(B-C) +cos(C-A) = -3/2 ,
find the value of c o s ( A ) + c o s ( B ) + c o s ( C ) cos(A) + cos(B) + cos(C) .


The answer is 0.

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Yash Jipkate by Yash Jipkate , 26, India

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1 solution

Yash Jipkate
Oct 20, 2014

Well its all about trigonometric identities cos(A-B) = cosAcosB + sinAsinB

therefore cos(A-B) + cos(B-C) + cos(C-A) = -3/2

cosAcosB + sinAsinB + cosBcosC + sinBsinC + cosCcosA + sinCsinA = -3/2

2(cosAcosB + sinAsinB + cosBcosC + sinBsinC + cosCcosA + sinCsinA) + 3 =0

2(cosAcosB + sinAsinB + cosBcosC + sinBsinC + cosCcosA + sinCsinA) + 1 +1+1 =0

2(cosAcosB + sinAsinB + cosBcosC + sinBsinC + cosCcosA + sinCsinA) + sin^{2}A + cos^{A} + sin^{B} +cos^{B} + sin^{C} + cos^{C} = 0

2cosAcosB + 2sinAsinB +2 cosBcosC + 2sinBsinC +2 cosCcosA + 2sinCsinA + sin^{2}A + cos^{A} + sin^{B} +cos^{B} + sin^{C} + cos^{C} = 0

[sin^{2}A +sin^{B} + sin^{C}+ 2sinAsinB + 2sinBsinC+ 2sinCsinA] + [cos^{2}A +cos^{B}+ cos^{C}+2cosAcosB +2 cosBcosC+2 cosCcosA] = 0

(sinA + sinB + sinC)^{2} + (cosA + cosB + cosC)^{2} = 0

Now since any number's square is always positive and sum of two can be zero only when both are zero.

cosA+ cosB + cosC = 0

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow, that's an amazing series of equations.

Calvin Lin Staff - 6 years, 7 months ago

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