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Minimize a certain function involving sine and cosine

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data
    It isn't a homework problem per se, but a curiosity a stumbled upon when trying to solve a physics problem (I was trying to calculate the angle I would need to do less work possible, while moving the box). The equation I found is:
    [itex]f(\theta)=\cos(\theta)+ 0.4sen(\theta)[/itex]


    2. Relevant equations

    Just the one stated above, and trig identities, probably.

    3. The attempt at a solution

    I tried a few things (including deriving and finding the roots of the function without success, since I couldn't found the roots), I also tried to rewrite (Using cos² + sin² = 1) and got:
    [itex]f(\theta)=\sqrt{1-\sin^2(\theta)}+ 0.4sin(\theta)[/itex]
    But I also don't know how to find the minimum in this equation.

    How could I go about solving that?

    Thanks in advance!
     
    Last edited: Jan 7, 2013
  2. jcsd
  3. Jan 7, 2013 #2
    What problems did you run into? Did you do anything with tangent?
     
  4. Jan 7, 2013 #3

    HallsofIvy

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    Interesting that you use both "sen" and "sin"! Can't decide between French and English?

    You have [itex]f(\theta)= cos(\theta)+ 0.4sin(\theta)[/itex] (only English for me, I'm afraid.). I don't see any reason to introduce a square root just to have only sine. Taking the derivative, [itex]f'(\theta)= -sin(\theta)+ 0.4cos(\theta)= 0[/itex] at a max or min. That is the same as [itex]sin(\theta)= 0.4 cos(\theta)[/itex] or, since sine and cosine are not 0 for the same [itex]\theta[/itex], we must have [itex]sin(\theta)/cos(\theta)= tan(\theta)= 0.4[/itex]. You can use a calculator to solve that.
     
    Last edited: Jan 7, 2013
  5. Jan 7, 2013 #4

    haruspex

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    Reposting HallsofIvy's post to fix up the LaTex:
     
  6. Jan 7, 2013 #5
    Oh god, I can't believe I ignored sin(x)/cos(x) = tan(x). I'm terribly sorry, thanks a lot! That solves my problem.

    About the whole sen and sin thing, it's because I'm actually Brazilian, and here we write "Sen", so I sometimes get both of them confused :P
     
  7. Jan 7, 2013 #6

    Ray Vickson

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    You also need to worry about whether the point you find is a maximizer or a minimizer of f.
     
  8. Jan 7, 2013 #7
    Yes indeed, but that was easily done by deriving again and finding if the result in this particular point would be positive or negative, since I found a negative value I concluded that it was a maximum, the value I wanted(In OP I miswrote, I actually wanted a maximum initially).

    [itex]tan(\theta)= 0.4;\theta = 21.8^o\\f''(21.8^o)= -1.077[/itex]

    However, since you mentioned that, I just realized that I have no idea on how to find the minimum. Shouldn't [itex]tan(\theta) = 0.4[\itex] yield two values, one of which is a maximum and another which is a minimum?
     
  9. Jan 7, 2013 #8

    Ray Vickson

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    Yes.
     
  10. Jan 7, 2013 #9
    Hmm. I found by trial and error that the angles should be: 21.8° and 201.8°, but how am I supposed to get the 201.8°? My calculator only gave me 21.8°.
     
  11. Jan 7, 2013 #10

    Dick

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    tan(x)=tan(x+180). The values of tan repeat every 180 degrees (pi in radians). It's periodic with period pi.
     
  12. Jan 7, 2013 #11
    Ohhh, I see, thanks!
    I really need to get better in trigonometry. I have several wrong concepts :S
     
  13. Jan 7, 2013 #12

    lurflurf

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    This is usually done with the identity

    [tex]\cos (x)+\frac{2}{5} \sin (x)=\sqrt {1+\left( \frac{2}{5} \right) ^2 } \sin \left( x+\arctan \left( \frac{5}{2} \right) \right)[/tex]

    Which is quite easy to optimize.
     
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