Given a Triangle with Angles a,b,cDate: 05/03/99 at 22:42:33 From: Tom Kellogg Subject: Trigonometry Given a triangle with angles a,b,c, show that cos(a)+cos(b)+cos(c) <= 3/2 This problem appears simple enough, yet it has caused me many hours of frustration. I've tried using the law of cosines, numerous trigonometric identities and several other approaches, but all seem to lead nowhere. Date: 05/12/99 at 03:50:29 From: Doctor Floor Subject: Re: Trigonometry Hi, Tom, Thanks for your question. For the solution we will need something we know about the graph of f(x) = cos(x). We know that this graph is convex on the interval [0, pi/2]. This means that if we have two unequal acute angles a and b, then the midpoint of (a,cos(a)) and (b,cos(b)) is below the graph. When one of the angles of triangle ABC is obtuse, say a > pi/2, then we know that b = pi-a-c, so b < pi-a. We can use this fact to show that for obtuse a and acute b, too, the midpoint of (a,cos(a)) and (b,cos(b)) is below the graph. To see this, note that the midpoint of (a,cos(a)) and (pi-a,cos(pi-a)) is (pi/2,0). Now we shift (pi-a, cos(pi-a)) to (b, cos(b)) = (pi-a-c,cos(b)). We find that the midpoint of (b, cos(b)) and (a,cos(a)) gets x-coordinate=(pi-c)/2. The y-coordinate of this midpoint has become higher by the shift; it becomes exactly [cos(b) - cos(pi-a)]/2. We now see, again using the convexity of cos(x) on [0,pi/2], that cos((pi-c)/2)= cos((pi-c)/2) - cos(pi/2) > cos(pi-a-c/2) - cos(pi-a) > [cos(pi-a-c) - cos(pi-a)]/2 = [cos(b) - cos(pi-a)]/2 = y-coordinate of midpoint (b, cos(b)) and (a,cos(a)) This shows that if a is obtuse, then the midpoint of (a,cos(a)) and (b,cos(b)) is below the graph of f(x) = cos(x), too. We can conclude, that if a,b,c are the unequal angles of a triangle, then the midpoints of (a,cos(a)), (b,cos(b)) and (c,cos(c)) are below the graph. This means that the centroid (the point of intersection of the medians) of the triangle formed by these three points is below the graph. This centroid can be calculated by taking the mean values of the x- and y-coordinates of the three points, hence (cos(a)+cos(b)+cos(c))/3 < cos((a+b+c)/3) = cos(pi/3) = 1/2 cos(a)+cos(b)+cos(c) < 3/2 This proves the inequality for a triangle with three unequal angles. If two of the three angles are equal, then one midpoint lies on the graph, but the centroid lies below the graph. And we have the same conclusion. When the triangle is equilateral, then it is easy to verify that cos(a)+cos(b)+cos(c) = 3/2. So we have proven the inequality. I hope this helped! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/