Question

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

Answer 1

Disprove?

Answer 2

A B n C r angles of a triangle

Answer 3

any one pleease help

Answer 4

ny hint ?

Answer 5

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 ?

Answer 6

hmn... cn we prove ny inequality for cos a + cos b + cos c

Answer 7

i like this question

Answer 8

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

Answer 9

@ashwinjohn3 That doesn't count as a proof. =/

Answer 10

@ParthKohli Then prove it by Mathematical induction.....

Answer 11

can you use the law of cosines /

Answer 12

i would argue that since this is symmetric in \(A,B,C\) the max must occur when \(A=B=C=\frac{\pi}{3}\)

Answer 13

but A , B n C r angles of a triangle.

Answer 14

How can each angle be pi/2 ? A + B + C = 180 @satellite73 sir

Answer 15

(its a three not a two)

Answer 16

Sorry :(

Answer 17

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

Answer 18

But can this hep us to proceed to get the proof ?

Answer 19

\[\cos(x) \le 1\]\[then \ \ \cos(90 - x) = \sin(x) \le 1\]\[also \ \ \cos(90) = 0\]Adding all,\[\cos(x) + \cos(90 - x) + \cos(90) \le 2\]BANG

Answer 20

\[A = x, B = 90 - x, C = 90\]

Answer 21

but this proof assumes that the triangle is right angled, which may not be the case.

Answer 22

i dont think we can assume a right angle

Answer 23

Oh, well.

Answer 24

yeah right we shall assume for any triangle

Answer 25

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

Answer 26

but we are also to prove for "equal to" condition

Answer 27

@mathslover \( n < r\) automatically satisfies \(n \le r\).

Answer 28

@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.

Answer 29

oh.. right