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 Acute triangle, Trigonometric function! (Posted on 2007-10-13)
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

 Subject Author Date Solution Bractals 2007-10-15 05:41:03 re(2): Solution K Sengupta 2007-10-14 10:59:29 re: proof (maybe) Charlie 2007-10-14 10:39:37 re: Solution Charlie 2007-10-14 10:37:43 Solution K Sengupta 2007-10-14 06:08:14 proof (maybe) Daniel 2007-10-14 05:32:49 Hint Bractals 2007-10-13 19:41:08 re: not a proof but ... Charlie 2007-10-13 16:47:30 not a proof but ... Charlie 2007-10-13 16:39:03

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