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ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Disprove?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
A B n C r angles of a triangle
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
any one pleease help
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
ny hint ?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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

mathslover Group TitleBest ResponseYou've already chosen the best response.0
hmn... cn we prove ny inequality for cos a + cos b + cos c
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i like this question
 one year ago

ashwinjohn3 Group TitleBest ResponseYou've already chosen the best response.0
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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
@ashwinjohn3 That doesn't count as a proof. =/
 one year ago

ashwinjohn3 Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli Then prove it by Mathematical induction.....
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
can you use the law of cosines /
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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

mathslover Group TitleBest ResponseYou've already chosen the best response.0
but A , B n C r angles of a triangle.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
How can each angle be pi/2 ? A + B + C = 180 @satellite73 sir
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
(its a three not a two)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Sorry :(
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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

mathslover Group TitleBest ResponseYou've already chosen the best response.0
But can this hep us to proceed to get the proof ?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
\[A = x, B = 90  x, C = 90\]
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.0
but this proof assumes that the triangle is right angled, which may not be the case.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i dont think we can assume a right angle
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
Oh, well.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
yeah right we shall assume for any triangle
 one year ago

FoolAroundMath Group TitleBest ResponseYou've already chosen the best response.0
cos(A) + cos(B) + cos(C) = \(\large \cos\frac{A+B}{2}\cos\frac{AB}{2}+\cos(C)\) = \(\large \sin\frac{C}{2}\cos\frac{AB}{2}+\cos C\) Let us fix angle C and try to maximise the sum. This will only happen when \(\large \cos\frac{AB}{2} = 1\) 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

mathslover Group TitleBest ResponseYou've already chosen the best response.0
but we are also to prove for "equal to" condition
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.2
@mathslover \( n < r\) automatically satisfies \(n \le r\).
 one year ago

FoolAroundMath Group TitleBest ResponseYou've already chosen the best response.0
@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

mathslover Group TitleBest ResponseYou've already chosen the best response.0
oh.. right
 one year ago
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