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Math Help - Find minimum value

  1. #1
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    Find minimum value

    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?
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  2. #2
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    Quote 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
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  3. #3
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    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
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  4. #4
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    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.
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  5. #5
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    Quote 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.
    Last edited by JakeD; July 11th 2006 at 03:43 PM.
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