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 need to find the minimum of $f(x)$ with $$f(x)=(\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x))^2$$ Could you help me with some clues?

share|cite|improve this question
Do you know what a derivative is? – ja72 Apr 13 '13 at 20:41
@ja72 yes, I know. – Queayiouer Apr 13 '13 at 20:45
I found the answer numerically, but I am not sure how to proceed analytically. – ja72 Apr 13 '13 at 20:48
Calculating the derivative will not work. – Lord Soth Apr 13 '13 at 20:49
@LordSoth it's true. Calculating the derivative the difficulty remains unchanged. – Queayiouer Apr 13 '13 at 20:51

You can do the following:

Let $$\sin x + \cos x = y$$

Then we have that $$\tan x + \cot x = \frac{2}{y^2 -1}$$


$$\sec x + \csc x = \frac{2y}{y^2-1}$$

I believe $$\sin x + \cos x + \tan x + \cot x + \sec x + \csc x $$ simplifies to $$y + \frac{2}{y-1}$$ (but I tried doing it in my head, so might have made mistakes).

And you can use the standard calculus techniques, now. (but take care to eliminate the corner cases etc).

share|cite|improve this answer
I can do it from here. Thank you. – Queayiouer Apr 13 '13 at 21:09
(+1) This is classic.:) My answer hint no where.. – Inceptio Apr 13 '13 at 21:11

To built upon @Queayiouer if you split the function as

$$ f(x) = \left( g(x) \right)^2 = \left( \left(\sin x+\frac{1}{\sin x}\right) + \left(\cos x+\frac{1}{\cos x}\right) + \left(\tan x+\frac{1}{\tan x}\right) \right)^2 $$

where the minimum occurs if $g(x)=0$ or $g'(x)=0$. The first is not going to happen.

So now we have

$$ g(x) = g_1(x) + g_2(x) + g_3(x) = \left(\sin x+\frac{1}{\sin x}\right) + \left(\cos x+\frac{1}{\cos x}\right) + \left(\tan x+\frac{1}{\tan x}\right) $$

Let find if they three functions have a common minimum since

$$ g_1'(x) = \cos(x) \left(1-\frac{1}{\sin^2 x} \right) = 0 $$ $$ g_2'(x) = \sin(x) \left(\frac{1}{\cos^2 x}-1 \right) = 0 $$ $$ g_3'(x) = \tan^2 x - \frac{1}{\tan^2 x} = 0$$

Unfortunately those are never all zero at the same time. So my guess it to proceed numerically with Newton-Raphson method where

$$ x \rightarrow x - \frac{ g_1(x) + g_2(x) + g_3(x) }{ g_1'(x) + g_2'(x) + g_3'(x) } $$

starting from $x=1$ or something. I get between $2.5<x<2.7$.

share|cite|improve this answer

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.