# Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$

I'm tutoring for a college math class and we're doing putnam problems next week and this one stumped me:

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

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Solutions to past Putnam exams are available in books and online. If this was a Putnam problem, what year? – Gerry Myerson Feb 13 '13 at 3:28
Where is the problem from? – Jonas Meyer Feb 13 '13 at 15:02
Found it, 2003 A3. Dan Bernstein's solution is at cr.yp.to/papers/putnam2003.ps and two or three solutions are at amc.maa.org/a-activities/a7-problems/putnam/-pdf/2003s.pdf – Gerry Myerson Feb 14 '13 at 0:21

Let $\sin{x}+\cos{x}=\sqrt{2}\sin{(x+\frac{\pi}{4})}=a$, then $a$ can take any value between $-\sqrt{2}$ and $\sqrt{2}$. We have $\sin{x}\cos{x}=\frac{a^2-1}{2}$. Thus

\begin{align} \left||\sin{x}+\cos{x}+\tan{x}+\cot{x}+\sec{x}+\csc{x} \right|& =\left|\sin{x}+\cos{x}+\frac{1}{\sin{x}\cos{x}}+\frac{\sin{x}+\cos{x}}{\sin{x}\cos{x}} \right| \\ & =\left|a+\frac{2+2a}{a^2-1} \right| \\ & =\left|a+\frac{2}{a-1} \right| \end{align}

If $a<1$, let $b=1-a>0$, then $a+\frac{2}{a-1}=1-(b+\frac{2}{b}) \leq 1-2\sqrt{2}<0$, so $\left|a+\frac{2}{a-1} \right| \geq 2\sqrt{2}-1$.

($b+\frac{2}{b} \geq 2\sqrt{2}$ by AM-GM inequality or square $\geq 0$)

Equality holds when $b=\sqrt{2}, a=1-\sqrt{2}, x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.

If $1 \leq a \leq \sqrt{2}$, then $a+\frac{2}{a-1}>\frac{2}{a-1}>2$, so $\left|a+\frac{2}{a-1} \right|>2>2\sqrt{2}-1$.

Thus the minimum value is $2\sqrt{2}-1$, achieved when $x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.

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Nice solution, very compact. Mine, although more straightforward, involved a lot of computation and reduced to a very simple result, a red flag that a simpler way is available. You should fix your LaTeX, though, as the functions look ugly. (+1) – Ron Gordon Feb 13 '13 at 11:12

Let $x \in [0,2 \pi)$ and

$$f(x) = \sin{x} + \cos{x}+\tan{x}+\cot{x}+\sec{x}+\csc{x}$$

Then

\begin{align}f'(x) &= \cos{x} - \sin{x} + \sec^2{x} - \csc^2{x} + \sec{x} \tan{x} - \csc{x} \cot{x}\\ &= \frac{\cos{x}-\sin{x}}{\cos^2{x} \sin^2{x}} (\cos^2{x} \sin^2{x} - (\cos{x} + \sin{x}) - (\cos^2{x}+\cos{x} \sin{x} + \sin^2{x}))\\ &= \frac{\cos{x}-\sin{x}}{\cos^2{x} \sin^2{x}} (\cos{x} \sin{x} - \cos{x} - \sin{x})(1 + \cos{x} \sin{x} + \cos{x}+\sin{x}) \end{align}

Regardless of the absolute value, we have extrema where

$$\begin{array} \\ \cos{x} - \sin{x} = 0 & (1) \\ \cos{x} \sin{x} - \cos{x} - \sin{x}=0 & (2) \\ 1 + \cos{x} \sin{x} + \cos{x}+\sin{x}=0 & (3) \end{array}$$

(1) implies that $x \in \{\pi/4,5 \pi/4\}$, while (3) implies that $x \in \{\pi,3 \pi/2\}$. (2) on the other hand implies that

$$\frac{1}{4} \sin^2{2 x} = 1 + \sin{2 x}$$

or that $2 x = 2 \pi-\arcsin{[2 (\sqrt{2}-1)]}$ so that the solution is $\in [0,2\pi)$. (The other solution to the quadratic is $> 1$.)

Plugging in these values into $g(x)$, we find that

\begin{align}\left |g \left ( \frac{\pi}{4} \right ) \right | &= 3 \sqrt{2} + 2\\ \left |g \left ( \frac{5\pi}{4} \right ) \right | &= 3 \sqrt{2} - 2\\ |g(\pi)| &= 2\\ \left |g \left ( \frac{3\pi}{2} \right ) \right | &=2\\\end{align}

Finally, for the last solution, I will spare you the simplifications involved; we get

$$\left|g \left ( \pi - \frac{1}{2} \arcsin{[2 (\sqrt{2}-1)]} \right ) \right | = 2 \sqrt{2}-1$$

The minimum value is thus $2 \sqrt{2}-1$.

EDIT

Per the point raised by @Ivan below, I show that $f(x) \ne 0 \: \forall x \in [0,2 \pi)$. $f(x)$ may be rewritten as

$$f(x) = \frac{\csc \left(\frac{\pi }{4}-\frac{x}{2}\right) \csc \left(\frac{x}{2}\right) (-\sin (x)-\cos (x)+2 \sin (x) \cos (x)+3)}{2 \sqrt{2}}$$

The solution to $f(x)=0$ means that

$$\sin^2{2 x} + 5 \sin{2 x} + 8 = 0$$

which has no real solutions. Thus, $f(x) \ne 0 \: \forall x \in [0,2 \pi)$ and the result holds.

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There is a minor flaw in this solution. If the function $f(x)$ can be equal to $0$ for some $x$, then the minimum value of $|f(x)|$ is $0$. However, $f(x)=0$ may not correspond to an extrema of $f(x)$. To give an example, suppose you want to minimise $|x^2-x|$ (Minimal value is 0). By your method, $f(x)=x^2-x$, then we have extrema when $f'(x)=2x-1=0$, i.e. when $x=\frac{1}{2}$. However $|f(\frac{1}{2})|=\frac{1}{4}$, which is not minimal. As such, you still need to check that $f(x)$ cannot be equal to $0$ for any $x$. – Ivan Loh Feb 13 '13 at 10:56
Aha, good point. I will add that. – Ron Gordon Feb 13 '13 at 11:01

I've found one solution here. Have a look.

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This link does not lead to a solution.... – Zilliput Feb 13 '13 at 15:04
I've updated the link. It seems to be working now. Have a look. I'm sorry for posting a redirecting link. – Ayush Khemka Feb 19 '13 at 12:08
Your updated link is exactly the link I put in the previous comment, 5 days ago. – Gerry Myerson Feb 19 '13 at 12:13
oh OK, I didn't pay attention to it probably. My bad. Sorry. – Ayush Khemka Feb 19 '13 at 12:21