Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x$.

share|cite|improve this question
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 and two or three solutions are at – 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 $x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.

share|cite|improve this answer
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

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


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

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 $.


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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.