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The following is a beautiful problem from Putnam 2003

minimize $|\sin x + \cos x + \tan x + \csc x + \sec x + \cot x|$

I was thinking about a small variation of the above problem

minimize $|\sin x + \cos x + \tan x - \csc x - \sec x - \cot x|$


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Both these values blow up certain trigonometric functions. So incorrect – user136833 Mar 20 '14 at 17:51
Is there a certain limit for $x$? $[0,\frac{\pi}{2}]$, for example? – 2012ssohn Mar 20 '14 at 17:52
@2012ssohn the original Putnam problem had no such limits. But then you naturally expect it to lie in $(-\pi, \pi)$ since the functions are periodic – user136833 Mar 20 '14 at 17:54
Can you represent each as complex exponentials then simplify? – Erik M Mar 20 '14 at 17:55
Ok. I was wondering about the original Putnam problem, and it seems to blow up at $x = 0$, and was wondering whether there are any limitations. – 2012ssohn Mar 20 '14 at 17:56
up vote 6 down vote accepted

Let $x=-\dfrac{\pi}{4}$. Then $\cos x = -\sin x, \sec x = -\csc x, \tan x = \cot x$ and the expression inside the absolute values is $0$.

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Beautiful. +1 for you. – MPW Mar 20 '14 at 17:58
+1 This is probably more beautiful than the original Putnam problem. – Yiyuan Lee Mar 20 '14 at 18:00
Adding integer multiples of $\pi$ gives you more zeros – David H Mar 20 '14 at 18:00
I believe the minimum value is 2, not 0? – Erik M Mar 20 '14 at 18:06
@ErikMiehling The problem to solve is the one in the gray box. The original problem is an old Putnam problem that inspired the new question. – John Habert Mar 20 '14 at 18:09

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