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Any triangle, Trigonometric function - second episode! (Posted on 2007-11-18) Difficulty: 3 of 5
Prove that in any triangle cos(A)+cos(B)+cos(C)>1

See The Solution Submitted by Chesca Ciprian    
Rating: 3.0000 (1 votes)

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non-obtuse case | Comment 1 of 12

In the interval 0<x<=pi/2, y = cos x is greater than or equal to y = -2x/pi + 1.

If, say, A<=B<=C are angles of a triangle in this interval, then

cos A > -2A/pi +1, cos B > -2B/pi +1, and cos C >= -2C/pi + 1

So summuing up the inequalities -->

cos A + cos B + cos C > 1.


  Posted by Dennis on 2007-11-18 15:21:40
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