## mathslover Group Title Prove tht cos A + cos B + cos C <= 2 one year ago one year ago

1. ParthKohli Group Title

Disprove?

2. mathslover Group Title

A B n C r angles of a triangle

3. mathslover Group Title

any one pleease help

4. mathslover Group Title

ny hint ?

5. shubhamsrg Group Title

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 Group Title

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

7. UnkleRhaukus Group Title

i like this question

8. ashwinjohn3 Group Title

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 Group Title

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

10. ashwinjohn3 Group Title

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

11. UnkleRhaukus Group Title

can you use the law of cosines /

12. satellite73 Group Title

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 Group Title

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

14. mathslover Group Title

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

15. UnkleRhaukus Group Title

(its a three not a two)

16. mathslover Group Title

Sorry :(

17. satellite73 Group Title

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 Group Title

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

19. ParthKohli Group Title

|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 Group Title

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

21. rajathsbhat Group Title

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

22. UnkleRhaukus Group Title

i dont think we can assume a right angle

23. ParthKohli Group Title

Oh, well.

24. mathslover Group Title

yeah right we shall assume for any triangle

25. FoolAroundMath Group Title

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 Group Title

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

27. ParthKohli Group Title

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

28. FoolAroundMath Group Title

@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 Group Title

oh.. right