Home > Just Math
Any triangle, Trigonometric function - second episode! (Posted on 2007-11-18) |
|
Prove that in any triangle
cos(A)+cos(B)+cos(C)>1
|
Submitted by Chesca Ciprian
|
Rating: 3.0000 (1 votes)
|
|
Solution:
|
(Hide)
|
Like in the first episode - see "Acute triangle, trigonometric function!" if we take on the graph of the function f(x) = cos(x) a point A(x,cos(x)) there will take shape between origin and and the point B(pi/2,1) 2 triangles and 1 trapezium. The area of the trapezium and one triangle is bigger than the area of the last triangle. After calculus we find that cos(x)>=1-2x/pi. Using this relation for the angle's of an acute triangle we find that cos(A)+cos(B)+cos(C) > 1 |
Comments: (
You must be logged in to post comments.)
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|