# Find minimum value

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• Jul 9th 2006, 01:28 PM
Jameson
Find minimum value
This is from the Putnam exam, 2003.

"Find the minimum value of $\displaystyle |\sin(x)+\cos(x)+\tan(x)+\cos(x)+\sec(x)+\csc(x)|$ for all real x".

My thoughts:

First I rewrote everything in terms of sine and cosine. After simplifying I get:

$\displaystyle \sin^2(x)\cos(x)+\sin(x)\cos^2(x)+\sin(x)\cos(x)+1$

Now here's my thinking... if I let a=sin(x) and b=cos(x) then I can use Lagrange Multipliers to find the minimum. What's the constraint though?
• Jul 9th 2006, 01:45 PM
CaptainBlack
Quote:

Originally Posted by Jameson
This is from the Putnam exam, 2003.

"Find the minimum value of $\displaystyle |\sin(x)+\cos(x)+\tan(x)+\cos(x)+\sec(x)+\csc(x)|$ for all real x".

My thoughts:

First I rewrote everything in terms of sine and cosine. After simplifying I get:

$\displaystyle \sin^2(x)\cos(x)+\sin(x)\cos^2(x)+\sin(x)\cos(x)+1$

Now here's my thinking... if I let a=sin(x) and b=cos(x) then I can use Lagrange Multipliers to find the minimum. What's the constraint though?

Should it be

$\displaystyle |\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x)|$?

RonL
• Jul 9th 2006, 02:59 PM
ThePerfectHacker
This problem with $\displaystyle y=|x|$ it is no diffrenciable at $\displaystyle x=0$. That was the problem with the line of best fit algorithm, unless you simplify the problem using the concept of least squares. If you procede as that then you need to minimize,
$\displaystyle y=(\cos x+\sin x+\csc x+\sec x+\cot x+\tan x)^2$
• Jul 11th 2006, 11:08 AM
Jameson
Semi-bump along with another solution.

Write f(x) as $\displaystyle \sin(x)+\cos(x)+\frac{1}{\sin(x)\cos(x)}$$\displaystyle +\frac{\sin(x)+\cos(x)}{\sin(x)\cos(x)} Now make the substitution \displaystyle \sin(x)+\cos(x)=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right) I don't follow this trig sub... any insight? I'll finish the posted solution after I get this step. • Jul 11th 2006, 02:19 PM JakeD Quote: Originally Posted by Jameson Semi-bump along with another solution. Write f(x) as \displaystyle \sin(x)+\cos(x)+\frac{1}{\sin(x)\cos(x)}$$\displaystyle +\frac{\sin(x)+\cos(x)}{\sin(x)\cos(x)}$

Now make the substitution $\displaystyle \sin(x)+\cos(x)=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)$

I don't follow this trig sub... any insight?

It follows from the identity

$\displaystyle \cos(y -x) = \sin(y)\sin(x) + \cos(y)\cos(x)$

where $\displaystyle y = \pi/4$ so $\displaystyle \sin(y) = \cos(y) = 1/\sqrt{2} .$

There is a general version

$\displaystyle a \sin(x) + b\cos(x) = \sqrt{a^2 + b^2} \cos(x - \tan^{-1}(a/b)) .$

Coincidence, I just learned this today from another site.