## mathslover Group Title cos A + cos B + cos C where A B and C are the angles of any triangle one year ago one year ago

1. hba

Kehna kia chate ho ?

2. hartnn

do we have options/choices for this ?

3. mathslover

iwant to simplify it...

4. hba

$\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }$

5. FoolAroundMath

$$\large = 1 + 4\sin{\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}$$

6. hartnn

cos A + cos B + cos C is in the most simplified form

7. hba

Yeah right :)

8. hartnn

in any triangle , A+B+C = $$\pi$$, so $$\cos(A+B)=-\cos C$$ $$\cos(C+B)=-\cos A$$ $$\cos(A+C)=-\cos B$$ if you want to complicate, i can help :P using cosine law.

9. mathslover

Well yes I wanted to use cosine law here and I got : cos A + cos B + cos C = $$\LARGE{-\frac{4s^3}{abc}}$$ where a , b and c are the sides of the triangle and s is semi perimeter of the triangle

10. mathslover

I did like this : $\large{\frac{(b^2+c^2-a^2)}{2bc} + \frac{(c^2+a^2-b^2)}{2ac} + \frac{(a^2+b^2-c^2)}{2ab}}$ and simpified this by taking LCM and taking terms common and I got : $\large{\frac{-4s^3}{abc}}$

11. hartnn

you mean u got $$ab^2+ac^2+ba^2+bc^2+ca^2+cb^2-a^3-b^3-c^3=-8s^3$$

12. mathslover

let me show you my complete work wait..

13. hartnn

that doesn't hold for an equilateral triangle, all sides equal, say =1

14. mathslover

Yes but I am getting this answer by simplifying cos A + cos B + cos C by using cosines law..

15. hartnn

ok, show your work, we'll find the error...

16. mathslover

ok 1 minute

17. mathslover

Oh! I think I got my mistake ... ..

18. mathslover

can you help me to prove that cos A + cos B + cos C is less than or equal to 2

19. mathslover

I got some conceptual answers previous time but I am searching for a numerical one.. @hartnn

20. ParthKohli

And mine was numerical but only for a right triangle =/

21. mathslover

=/ yeah @hartnn any try?

22. hartnn

if i get something, i'll post.

23. mathslover

${cos A + cos B + cos C = \\ \frac{a(b^2+c^2) + b(a^2+c^2) + \\c(a^2+b^2) - (a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{2abc}}$

24. AravindG

trying on paper

25. mathslover

Thanks! I am also trying my best...

26. AravindG

got it !

27. mathslover

?

28. AravindG

|dw:1356616135337:dw|

29. mathslover

yes..

30. AravindG

now cos A+cos B+Cos C is always less than equal to 3

31. mathslover

We have to prove for less than or equal to 2

32. AravindG

ya i am proving

33. mathslover

ok

34. AravindG

|dw:1356616236014:dw|

35. AravindG

you get that >?

36. mathslover

No :(

37. AravindG

why not ?where u have doubt ?

38. mathslover

cos A + cos B - cos(A+B) , how is it less than or equal to 3 ?

39. AravindG

i wrote cos A+cos B+cos(18-(A+B)<=3

40. AravindG

this is because maximum value of cos is 1

41. AravindG

ie 1+1+1=3

42. AravindG

all other cases cos a +cos b +cos c value will be less than 3

43. AravindG

and note i wrote c as 180-(A+B)

44. AravindG

@mathslover you get it?

45. mathslover

wait!

46. AravindG

47. mathslover

Are you taking A = B = 0 degrees?

48. mathslover

since cos A + cos B + cos (180-(A+B)) = 3 is not possible ?

49. mathslover

I may be wrong but please correct me if I am ...

50. AravindG

yep for maximum value we take A=B=C=0 degree

51. mathslover

but A , B and C are the angles of a triangle..

52. AravindG

yess that was in general case

53. AravindG

in the triangle case angle c is replaced by 180-(A+B)

54. AravindG

hence cos C becomes cos (180-(A+B))

55. mathslover

Yes but we can not take the angles of a triangle as 0 degrees..

56. AravindG

i didnt told you to take !!i said that is general case!

57. AravindG

but equal to not possible

58. AravindG

in triangle you have equation cos A+cos B+cos(180-(A+B))<3

59. mathslover

so can cos A + cos B +cos C be greater than 2 also ?

60. AravindG

no ! thats where we apply trigonometry ! os(180-x)=-sin x

61. AravindG

you understand that much?

62. mathslover

cos ( 180 - x ) = - cos (x)

63. AravindG

oops srry ya

64. AravindG

-cos x

65. AravindG

so our equation for triangle becomes cos A+cos B-cos(A+B)<3

66. mathslover

got it now thanks a lot

67. AravindG

:) you are welcome bro!

68. hartnn

directly comes from,$$\cos(A+B)=-\cos C$$ than from cos A+cos B-cos(A+B)<3 how u proved cos A+cos B+cos C<=2

69. hartnn

?

70. AravindG

simple cos A +cos B maximum can be 2 and -cos(a+b) is a reducing term so it always results in value less than or equal to 2 for the expression

71. hartnn

cos can be negative

72. hartnn

ohh..in 1st and 4th Quadrant cos is positive....

73. AravindG

ya whatever value cos takes it wont exceed +2

74. AravindG

yep !

75. hartnn

i thought cos(A+B) could be negative, thats why i doubted..but cos (A+B) will be positive...

76. AravindG

;)

77. AravindG