Putnam Optimization Hard Problem

Find the minimum of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$.

This is a horrible problem, you could have wasted an inordinate amount of your precious time finding how to, for example, express this given expression in terms of only one variable instead of the given six and then work from there because the methods that could be employed are not necessarily immediately obvious.  But that's life. If you want to acquire the skills of thinking flexibly and creatively, you need to solve complex and unfamiliar problems.

If you are a mathematics educator, then this kind of problem is what we should shower our students because we must acknowledge that the stronger performance of our students, the more it reflects our curricular emphasis and our schools’ efforts in developing problem solving skills in our students.

Now, back to our problem. There are at least two ways to tackle this problem effectively and I will lead you one after another on two separate blog posts:

Method I:

If we let $x=a-135^{\circ}$, where $a\in R$, we see that we have:

$\begin{align*}\sin x&=\sin (a-135^{\circ})\\&=\dfrac{-\sin a-\cos a}{\sqrt{2}}\end{align*}$ and

$\begin{align*}\cos x&=\cos (a-135^{\circ})\\&=\dfrac{-\cos a+\sin a}{\sqrt{2}}\end{align*}$ and adding these two yields

$\begin{align*}\sin x+\cos x&=\dfrac{-\sin a-\cos a}{\sqrt{2}}+\dfrac{-\cos a+\sin a}{\sqrt{2}}\\&=-\sqrt{2}\cos a\end{align*}$

Similarly,

$\begin{align*}\tan x&=\tan(a-135^{\circ})\\&=\dfrac{\sin a+\cos a}{\cos a-\sin a}\end{align*}$ and

$\begin{align*}\cot x&=\cot(a-135^{\circ})\\&=\dfrac{\cos a-\sin a}{\sin a+\cos a}\end{align*}$ and adding them up gives

$\begin{align*}\tan x+\cot x&=\dfrac{\sin a+\cos a}{\cos a-\sin a}+\dfrac{\cos a-\sin a}{\sin a+\cos a}\\&=\dfrac{2}{\cos^2 a-\sin^2 a}\end{align*}$

Finally,

$\begin{align*}\sec x&=\dfrac{1}{\cos x}\\&=-\dfrac{\sqrt{2}}{\sin a+\cos a}\end{align*}$ and

$\begin{align*}\csc x&=\dfrac{1}{\sin x}\\&=\dfrac{\sqrt{2}}{\sin a-\cos a}\end{align*}$, adding the two gives

$\begin{align*}\sec x+\csc x&=-\dfrac{\sqrt{2}}{\sin a+\cos a}+\dfrac{\sqrt{2}}{\sin a-\cos a}\\&=\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}\end{align*}$

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

$=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}+\dfrac{2\sqrt{2}\cos a}{\sin^2 a-\cos^2 a}|$

$=|-\sqrt{2}\cos a+\dfrac{2}{\cos^2 a-\sin^2 a}-\dfrac{2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|$

$=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{\cos^2 a-\sin^2 a}|$

$=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{2\cos^2 a-1}|$

$=|-\sqrt{2}\cos a+\dfrac{2-2\sqrt{2}\cos a}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|$

$=|-\sqrt{2}\cos a+\dfrac{2(1-\sqrt{2}\cos a)}{(\sqrt{2}\cos a-1)(\sqrt{2}\cos a+1)}|$

$=|-\sqrt{2}\cos a-\dfrac{2}{\sqrt{2}\cos a+1}|$

$=|-\left(\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} \right)|$

$=|\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1}|$

$=|\sqrt{2}\cos a+1+\dfrac{2}{\sqrt{2}\cos a+1}-1|$

Treat $\sqrt{2}\cos a>0$, we can hence apply the AM-GM inequality to the following:

$(\sqrt{2}\cos a+1)+\dfrac{2}{\sqrt{2}\cos a+1}\ge 2\sqrt{\cancel{(\sqrt{2}\cos a+1)}\left(\dfrac{2}{\cancel{(\sqrt{2}\cos a+1)}} \right)}= 2\sqrt{2}$

Since $\sqrt{2}\cos a+\dfrac{2}{\sqrt{2}\cos a+1} = 2 \sqrt{2}-1$ does have solutions, we can conclude by now that

$\displaystyle |\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ is $2 \sqrt{2}-1$.