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

1. ParthKohli

Disprove?

2. mathslover

A B n C r angles of a triangle

3. mathslover

any one pleease help

4. mathslover

ny hint ?

5. 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 ?

6. mathslover

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

7. UnkleRhaukus

i like this question

8. 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

9. ParthKohli

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

10. anonymous

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

11. UnkleRhaukus

can you use the law of cosines /

12. anonymous

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

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

14. mathslover

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

15. UnkleRhaukus

(its a three not a two)

16. mathslover

Sorry :(

17. 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

18. mathslover

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

19. 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

20. ParthKohli

$A = x, B = 90 - x, C = 90$

21. anonymous

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

22. UnkleRhaukus

i dont think we can assume a right angle

23. ParthKohli

Oh, well.

24. mathslover

yeah right we shall assume for any triangle

25. 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

26. mathslover

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

27. ParthKohli

@mathslover $$n < r$$ automatically satisfies $$n \le r$$.

28. 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.

29. mathslover

oh.. right