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Acute triangle, Trigonometric function! (Posted on 2007-10-13) Difficulty: 3 of 5
Prove that in an acute triangle sin(A) + sin(B) + sin(C) > 2

  Submitted by Chesca Ciprian    
Rating: 4.0000 (1 votes)
Solution: (Hide)
From the graph of the sin(x) function between 0 and PI/2 if we take 3 points A(x, sin(x)), origin and (PI/2,1) there will take shape 1 triangle and otherwise 1 triangle and 1 trapezium. If we consider that the area of the first triangle is smaller then the sum of the second triangle and the trapezium we will find the relation (well know!!!) sin(x)>=2*x/PI. After we replace in this relation the angle A,B, and C and sum this 3 relation we will find sin(A) + sin(B) + sin(C) > 2

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionBractals2007-10-15 05:41:03
re(2): SolutionK Sengupta2007-10-14 10:59:29
re: proof (maybe)Charlie2007-10-14 10:39:37
Questionre: SolutionCharlie2007-10-14 10:37:43
SolutionSolutionK Sengupta2007-10-14 06:08:14
proof (maybe)Daniel2007-10-14 05:32:49
Hints/TipsHintBractals2007-10-13 19:41:08
Some Thoughtsre: not a proof but ...Charlie2007-10-13 16:47:30
Some Thoughtsnot a proof but ...Charlie2007-10-13 16:39:03
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