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mathslover

Prove tht cos A + cos B + cos C <= 2

  • one year ago
  • one year ago

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  1. ParthKohli
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    Disprove?

    • one year ago
  2. mathslover
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    A B n C r angles of a triangle

    • one year ago
  3. mathslover
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    any one pleease help

    • one year ago
  4. mathslover
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    ny hint ?

    • one year ago
  5. shubhamsrg
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    is it ever equal to 2 ? i'd say the max occurs when a=b=c i.e. max value is 3/2 ? isnt it ? you can verify from hit and trial ?

    • one year ago
  6. mathslover
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    hmn... cn we prove ny inequality for cos a + cos b + cos c

    • one year ago
  7. UnkleRhaukus
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    i like this question

    • one year ago
  8. ashwinjohn3
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    Inputting A=60,B=90,C=30 cos 60+cos 90+cos 30=1/2+0+[(3)^1/2]/2=[1+(3)^1/2]/2=(1+1.7)/2=2.7/2=1.35 which is less than 2

    • one year ago
  9. ParthKohli
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    @ashwinjohn3 That doesn't count as a proof. =/

    • one year ago
  10. ashwinjohn3
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    @ParthKohli Then prove it by Mathematical induction.....

    • one year ago
  11. UnkleRhaukus
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    can you use the law of cosines /

    • one year ago
  12. satellite73
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    i would argue that since this is symmetric in \(A,B,C\) the max must occur when \(A=B=C=\frac{\pi}{3}\)

    • one year ago
  13. mathslover
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    but A , B n C r angles of a triangle.

    • one year ago
  14. mathslover
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    How can each angle be pi/2 ? A + B + C = 180 @satellite73 sir

    • one year ago
  15. UnkleRhaukus
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    (its a three not a two)

    • one year ago
  16. mathslover
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    Sorry :(

    • one year ago
  17. satellite73
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    yea, the same triangle and therefore you cannot tell \(A,B,C\) apart, which is what i meant when i said it is symmetric. you label the triangle one way, i label it another, we have the same thing which is why the max occurs when they are equal

    • one year ago
  18. mathslover
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    But can this hep us to proceed to get the proof ?

    • one year ago
  19. ParthKohli
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    |dw:1356530855111:dw|\[\cos(x) \le 1\]\[then \ \ \cos(90 - x) = \sin(x) \le 1\]\[also \ \ \cos(90) = 0\]Adding all,\[\cos(x) + \cos(90 - x) + \cos(90) \le 2\]BANG

    • one year ago
  20. ParthKohli
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    \[A = x, B = 90 - x, C = 90\]

    • one year ago
  21. rajathsbhat
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    but this proof assumes that the triangle is right angled, which may not be the case.

    • one year ago
  22. UnkleRhaukus
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    i dont think we can assume a right angle

    • one year ago
  23. ParthKohli
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    Oh, well.

    • one year ago
  24. mathslover
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    yeah right we shall assume for any triangle

    • one year ago
  25. FoolAroundMath
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    cos(A) + cos(B) + cos(C) = \(\large \cos\frac{A+B}{2}\cos\frac{A-B}{2}+\cos(C)\) = \(\large \sin\frac{C}{2}\cos\frac{A-B}{2}+\cos C\) Let us fix angle C and try to maximise the sum. This will only happen when \(\large \cos\frac{A-B}{2} = 1\) i.e. \(A=B\). Similarly, if we fix B and maximise the sum, then A=C fix A => sum is maximum when B=C so as long as any two angles are unequal, the sum can be increased further. => that the maximum occurs when all three are equal. The sum cant be maximised further when A=B=C. So the max. occurs when A=B=C and the max value = 3/2 <2

    • one year ago
  26. mathslover
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    but we are also to prove for "equal to" condition

    • one year ago
  27. ParthKohli
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    @mathslover \( n < r\) automatically satisfies \(n \le r\).

    • one year ago
  28. FoolAroundMath
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    @mathslover indirectly, what I am trying to do here is to show that, if any two angles are unequal, the sum can be increased further. So, if we were trying to find the max value, we would wish to make all pairs of angles equal so that the sum cannot be further maximised and will have attained its maximum value.

    • one year ago
  29. mathslover
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    oh.. right

    • one year ago
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