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mathslover

  • 2 years ago

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

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  1. ParthKohli
    • 2 years ago
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    Disprove?

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

  3. mathslover
    • 2 years ago
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    any one pleease help

  4. mathslover
    • 2 years ago
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    ny hint ?

  5. shubhamsrg
    • 2 years 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
    • 2 years ago
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    hmn... cn we prove ny inequality for cos a + cos b + cos c

  7. UnkleRhaukus
    • 2 years ago
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    i like this question

  8. ashwinjohn3
    • 2 years 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
    • 2 years ago
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    @ashwinjohn3 That doesn't count as a proof. =/

  10. ashwinjohn3
    • 2 years ago
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    @ParthKohli Then prove it by Mathematical induction.....

  11. UnkleRhaukus
    • 2 years ago
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    can you use the law of cosines /

  12. satellite73
    • 2 years 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
    • 2 years ago
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    but A , B n C r angles of a triangle.

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

  15. UnkleRhaukus
    • 2 years ago
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    (its a three not a two)

  16. mathslover
    • 2 years ago
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    Sorry :(

  17. satellite73
    • 2 years 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
    • 2 years ago
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    But can this hep us to proceed to get the proof ?

  19. ParthKohli
    • 2 years 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
    • 2 years ago
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    \[A = x, B = 90 - x, C = 90\]

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

  22. UnkleRhaukus
    • 2 years ago
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    i dont think we can assume a right angle

  23. ParthKohli
    • 2 years ago
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    Oh, well.

  24. mathslover
    • 2 years ago
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    yeah right we shall assume for any triangle

  25. FoolAroundMath
    • 2 years 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
    • 2 years ago
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    but we are also to prove for "equal to" condition

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

  28. FoolAroundMath
    • 2 years 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
    • 2 years ago
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    oh.. right

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