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Cosines everywhere (Posted on 2005-03-21) Difficulty: 4 of 5
Prove that in any triangle ABC, 8.cos(A).cos(B).cos(C) < 1.

  Submitted by Federico Kereki    
Rating: 4.5000 (2 votes)
Solution: (Hide)
We have 1-8.cos(A).cos(B).cos(C) =

Since cos(B+C)= -cos(π-B-C)= -cos(A), the expression above is:
1-4.cos(A).cos(B-C) + 4.cos(A)² =
sin(B-C)² + cos(B-C)² -4.cos(A).cos(B-C) + 4.cos(A)² =
sin(B-C)² + [cos(B-C)-2cos(A)]² which is always non negative.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
probable mistakeJohn2006-10-16 01:18:32
re(2): This problem needs some formatting changes...Erik O.2005-04-05 22:38:46
Advanced Calculus ProblemRichard2005-03-23 22:55:31
re: This problem needs some formatting changes...Richard2005-03-23 03:07:33
This problem needs some formatting changes...Erik O.2005-03-22 17:25:46
SolutionSolutionDavid Shin2005-03-21 19:13:57
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