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 Trigonometry Trigonometry Math Forum

 March 10th, 2008, 10:18 AM #1 Newbie   Joined: Feb 2008 Posts: 6 Thanks: 0 Inequality with cosine Prove that 1/(cos(A)+1/cos(B)+1/cos(C)>=6 if 0
 June 6th, 2008, 03:44 PM #2 Newbie   Joined: Jun 2008 From: New Zealand Posts: 20 Thanks: 0 We need to use the following inequalities: (1) cosA + cosB + cosC <= 3/2 for more tetails about this inequality, visit http://mathcircle.berkeley.edu/archi...espdf/trig.pdf (2) (x + y + z)/3 >= cube root of xyz (AM-GM Inequality) So cube root of cosAcosBcosC <= (cosA + cosB + cosC)/3 <= (3/2)/3 = 1/2 Now 1/cosA + 1/cosB + 1/cosC <= 3*cube root of [1/(cosAcosBcosC)] = 3/ cube root of (cosAcosBcosC) >= 3/(1/2) = 6 Thus 1/cosA + 1/cosB + 1/cosC >= 6 June 7th, 2008, 05:44 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Alternative way to start: 0 ? sin�(B - C) + (cos(B - C) - 2cos A)� 0 ? sin�(B - C) + cos�(B - C) - 4cos A(cos(B - C) - cos A) but cos(B - C) - cos A = cos(B - C) + cos(B + C) = 2cos(B)cos(C), so 0 ? 1 - 8cos(A)cos(B)cos(C), and so the cube root of cos(A)cos(B)cos(C) ? 1/2 if A, B and C are acute angles. Tags cosine, inequality ,

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# cosine inequality

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