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Sinx+cosx> = 1

  1. Apr 29, 2010 #1
    Is there any non geometrical way to proof this fact?
    sinx+cosx>=1 for every x in [0,Pi/2]
     
    Last edited: Apr 29, 2010
  2. jcsd
  3. Apr 29, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi estro! Welcome to PF! :smile:
    i] it's not true for 90º < x < 360º

    ii] you can prove non-geometrically that sinx + cosx = sin(x + 45º)/sin45º :wink:
     
  4. Apr 29, 2010 #3
    Re: Welcome to PF!

    Oh, sorry I corrected my mistake.
     
  5. Apr 30, 2010 #4
    Re: sinx+cosx>=1

    [tex]f(x) = sinx + cosx[/tex]

    [tex]f'(x) = cosx - sinx = 0[/tex]

    [tex]cosx = sinx[/tex]

    [tex]x = \frac{\pi}{4}\\[/tex]
    For values < pi/4, f'(x) = +, and for values >pi/4, f'(x) = -. This means at x=pi/4, there is a local maximum on the interval [0,pi/2]. Therefore, x=0 and x=pi/2 must be local minimums. We can then say that for all values x on the interval [0,pi/2]

    [tex]f(0 or \frac{\pi}{2}) \le f(x) \le f(\frac{\pi}{4})\\[/tex]
    [tex]1 \le f(x) \le \sqrt{2}\\[/tex]

    I sort of just made this up off the spot.
     
  6. Apr 30, 2010 #5

    nicksauce

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    Re: sinx+cosx>=1

    Well if you square both sides it's just

    sin^2 + cos^x + 2sinxcosx >= 1

    2sinxcosx >= 0

    Which is obviously true in the first quadrant.
     
  7. Apr 30, 2010 #6
    Re: sinx+cosx>=1

    I think from obvious reasons your idea is wrong...
     
  8. Apr 30, 2010 #7

    mathman

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    Re: sinx+cosx>=1

    Since 1 ≥ (sinx, cosx) ≥ 0 in the interval, sinx ≥ sin2x and cosx ≥ cos2x.
    Therefore sinx + cosx ≥ sin2x + cos2x = 1.
     
    Last edited: Apr 30, 2010
  9. Apr 30, 2010 #8

    D H

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    Re: sinx+cosx>=1

    What obvious reason? nicksauce's argument is correct regarding the first quadrant.

    That said, his argument is not correct throughout. [itex]\sin x \cos x \ge 0[/itex] for the third quadrant as well. In that quadrant, however, [itex]\sin x + \cos x \le -1[/itex].
     
  10. Apr 30, 2010 #9
    Re: sinx+cosx>=1

    From nicksauce's argument, we can't conclude sinx+cosx >=1 for x in [0,Pi/2].
     
  11. Apr 30, 2010 #10

    LCKurtz

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    Re: sinx+cosx>=1

    You can get both from nick's argument. For x in quadrant I or III:

    2 sin x cos x ≥ 0
    sin2(x) + 2 sin(x)cos(x) + cos2x ≥ 1
    ((sin(x) + cos(x))2 ≥ 1
    |sin(x) + cos(x)| ≥ 1

    Both results follow considering the signs in those two quadrants.
     
  12. Apr 30, 2010 #11
    Re: sinx+cosx>=1

    Is nick's argument weak because he assumed the premiss is true, then deduced his conclusion on that assumption?
     
  13. May 1, 2010 #12
    Re: sinx+cosx>=1

    That's what I was thinking: that you can't subtract 1 from or square both sides unless the equation's true in the first place.
     
  14. May 1, 2010 #13

    rock.freak667

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    Re: sinx+cosx>=1

    I'd think to write sinx+cosx in the form Rsin(x+β) and then show what happens in the range [0,π/2]
     
  15. May 1, 2010 #14

    tiny-tim

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    Why is almost everyone questioning nicksauce's :smile: proof?

    It's perfectly valid in the first quadrant ([0,π/2]), which is all the original question asks for!! :rolleyes:

    (and it's simpler than my proof!)

    (and as LCKurtz :smile: says, it can be adapted to cover the whole circle)
    see my first post :wink:
     
    Last edited: May 1, 2010
  16. May 1, 2010 #15

    nicksauce

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    Re: sinx+cosx>=1

    True. I wasn't trying to provide a rigorous proof, but rather roughly sketching out the means for a direction by which the OP could arrive at a rigorous proof.
     
  17. May 1, 2010 #16

    D H

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    Re: sinx+cosx>=1

    Which is a good thing. We aren't supposed to solve the students problems for them here. We (me included) came a little too close to that in this thread.

    nicksauce's argument is perfectly valid given the right circumstances. The OP needs to show that these circumstances apply here.
     
  18. May 3, 2010 #17
    Re: sinx+cosx>=1

    Sorry if someone else presented the same argument in a way i couldn't understand:

    (x+y)^2 >= x^2+y^2

    so x+y >= (x^2+y^2)^(1/2)

    and so sin(x)+cos(x) >= (sin^2(x)+cos^2(x))^(1/2)=1
    so sin(x)+cos(x) >= 1

    or do you consider the triangle inequality "geometric"? (which, by the name of "triangle", i would not begrudge you for :P)
     
  19. May 3, 2010 #18

    Char. Limit

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    Re: sinx+cosx>=1

    Well, the first part of the equation is only true if...

    [tex](x+y)^2 \geq x^2 + y^2[/tex]

    [tex]x^2+2xy+y^2 \geq x^2+y^2[/tex]

    [tex]2xy \geq 0[/tex]

    So sin(x)+cos(x) >= 1 only holds true if both sin(x) and cos(x) are positive, I suppose.
     
  20. May 3, 2010 #19

    D H

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    Re: sinx+cosx>=1

    Exactly.
     
  21. May 3, 2010 #20
    Re: sinx+cosx>=1

    I hope you understand that I've already solved this problem before posting this question on PF.
    And I never asked for a formal proof, I indeed asked for an alternative idea to what I've used in my geometrical based proof.
    And still I couldn't figure out pure "calculus" approach to solving this problem.

    It's a little pity that this thread was flooded with unclear algebra rather ideas...

    Anyway thank you all very much...
     
    Last edited: May 3, 2010
  22. May 3, 2010 #21

    LCKurtz

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    Re: sinx+cosx>=1

    It's a pity??? It looks to me like you received several alternative ideas.
     
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