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mathslover

  • one year ago

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

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  1. ParthKohli
    • one year ago
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    Disprove?

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

  3. mathslover
    • one year ago
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    any one pleease help

  4. mathslover
    • one year ago
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    ny hint ?

  5. shubhamsrg
    • one year ago
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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 ?

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

  7. UnkleRhaukus
    • one year ago
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    i like this question

  8. ashwinjohn3
    • one year ago
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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

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

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

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

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

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

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

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

  16. mathslover
    • one year ago
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    Sorry :(

  17. satellite73
    • one year ago
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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

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

  19. ParthKohli
    • one year ago
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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

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

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

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

  23. ParthKohli
    • one year ago
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    Oh, well.

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

  25. FoolAroundMath
    • one year ago
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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

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

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

  28. FoolAroundMath
    • one year ago
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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.

  29. mathslover
    • one year ago
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    oh.. right

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