Find minimum value

**Jameson** · July 9th 2006, 01:28 PM

This is from the Putnam exam, 2003.

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

My thoughts:

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

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

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?

**CaptainBlack** · July 9th 2006, 01:45 PM

Originally Posted by Jameson

This is from the Putnam exam, 2003.

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

My thoughts:

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

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

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

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

RonL

**ThePerfectHacker** · July 9th 2006, 02:59 PM

This problem with $y=|x|$ it is no diffrenciable at $x=0$ . 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,
$y=(\cos x+\sin x+\csc x+\sec x+\cot x+\tan x)^2$

**Jameson** · July 11th 2006, 11:08 AM

Semi-bump along with another solution.

Write f(x) as $\sin(x)+\cos(x)+\frac{1}{\sin(x)\cos(x)}$ $+\frac{\sin(x)+\cos(x)}{\sin(x)\cos(x)}$

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

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

I'll finish the posted solution after I get this step.

**JakeD** · July 11th 2006, 02:19 PM

Originally Posted by Jameson

Semi-bump along with another solution.

Write f(x) as $\sin(x)+\cos(x)+\frac{1}{\sin(x)\cos(x)}$ $+\frac{\sin(x)+\cos(x)}{\sin(x)\cos(x)}$

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

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

It follows from the identity

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

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

There is a general version

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

Coincidence, I just learned this today from another site.

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