mathslover
  • mathslover
Prove tht cos A + cos B + cos C <= 2
Mathematics
schrodinger
  • schrodinger
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ParthKohli
  • ParthKohli
Disprove?
mathslover
  • mathslover
A B n C r angles of a triangle
mathslover
  • mathslover
any one pleease help

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mathslover
  • mathslover
ny hint ?
shubhamsrg
  • shubhamsrg
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 ?
mathslover
  • mathslover
hmn... cn we prove ny inequality for cos a + cos b + cos c
UnkleRhaukus
  • UnkleRhaukus
i like this question
anonymous
  • anonymous
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
ParthKohli
  • ParthKohli
@ashwinjohn3 That doesn't count as a proof. =/
anonymous
  • anonymous
@ParthKohli Then prove it by Mathematical induction.....
UnkleRhaukus
  • UnkleRhaukus
can you use the law of cosines /
anonymous
  • anonymous
i would argue that since this is symmetric in \(A,B,C\) the max must occur when \(A=B=C=\frac{\pi}{3}\)
mathslover
  • mathslover
but A , B n C r angles of a triangle.
mathslover
  • mathslover
How can each angle be pi/2 ? A + B + C = 180 @satellite73 sir
UnkleRhaukus
  • UnkleRhaukus
(its a three not a two)
mathslover
  • mathslover
Sorry :(
anonymous
  • anonymous
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
mathslover
  • mathslover
But can this hep us to proceed to get the proof ?
ParthKohli
  • ParthKohli
|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
ParthKohli
  • ParthKohli
\[A = x, B = 90 - x, C = 90\]
anonymous
  • anonymous
but this proof assumes that the triangle is right angled, which may not be the case.
UnkleRhaukus
  • UnkleRhaukus
i dont think we can assume a right angle
ParthKohli
  • ParthKohli
Oh, well.
mathslover
  • mathslover
yeah right we shall assume for any triangle
FoolAroundMath
  • FoolAroundMath
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
mathslover
  • mathslover
but we are also to prove for "equal to" condition
ParthKohli
  • ParthKohli
@mathslover \( n < r\) automatically satisfies \(n \le r\).
FoolAroundMath
  • FoolAroundMath
@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.
mathslover
  • mathslover
oh.. right

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