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mathslover
 3 years ago
Prove tht cos A + cos B + cos C <= 2
mathslover
 3 years ago
Prove tht cos A + cos B + cos C <= 2

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mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0A B n C r angles of a triangle

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0any one pleease help

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0is 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
 3 years ago
Best ResponseYou've already chosen the best response.0hmn... cn we prove ny inequality for cos a + cos b + cos c

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1i like this question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Inputting 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
 3 years ago
Best ResponseYou've already chosen the best response.2@ashwinjohn3 That doesn't count as a proof. =/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli Then prove it by Mathematical induction.....

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1can you use the law of cosines /

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i would argue that since this is symmetric in \(A,B,C\) the max must occur when \(A=B=C=\frac{\pi}{3}\)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0but A , B n C r angles of a triangle.

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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1(its a three not a two)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea, 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
 3 years ago
Best ResponseYou've already chosen the best response.0But can this hep us to proceed to get the proof ?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2dw: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
 3 years ago
Best ResponseYou've already chosen the best response.2\[A = x, B = 90  x, C = 90\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but this proof assumes that the triangle is right angled, which may not be the case.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1i dont think we can assume a right angle

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0yeah right we shall assume for any triangle

FoolAroundMath
 3 years ago
Best ResponseYou've already chosen the best response.0cos(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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0but we are also to prove for "equal to" condition

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2@mathslover \( n < r\) automatically satisfies \(n \le r\).

FoolAroundMath
 3 years ago
Best 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.
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