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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

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  1. hba Group Title
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    Kehna kia chate ho ?

    • one year ago
  2. hartnn Group Title
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    do we have options/choices for this ?

    • one year ago
  3. mathslover Group Title
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    iwant to simplify it...

    • one year ago
  4. hba Group Title
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    \[\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }\]

    • one year ago
  5. FoolAroundMath Group Title
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    \(\large = 1 + 4\sin{\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}\)

    • one year ago
  6. hartnn Group Title
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    cos A + cos B + cos C is in the most simplified form

    • one year ago
  7. hba Group Title
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    Yeah right :)

    • one year ago
  8. hartnn Group Title
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    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.

    • one year ago
  9. mathslover Group Title
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    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

    • one year ago
  10. mathslover Group Title
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    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 : \[\large{\frac{-4s^3}{abc}}\]

    • one year ago
  11. hartnn Group Title
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    you mean u got \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2-a^3-b^3-c^3=-8s^3\)

    • one year ago
  12. mathslover Group Title
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    let me show you my complete work wait..

    • one year ago
  13. hartnn Group Title
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    that doesn't hold for an equilateral triangle, all sides equal, say =1

    • one year ago
  14. mathslover Group Title
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    Yes but I am getting this answer by simplifying cos A + cos B + cos C by using cosines law..

    • one year ago
  15. hartnn Group Title
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    ok, show your work, we'll find the error...

    • one year ago
  16. mathslover Group Title
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    ok 1 minute

    • one year ago
  17. mathslover Group Title
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    Oh! I think I got my mistake ... ..

    • one year ago
  18. mathslover Group Title
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    can you help me to prove that cos A + cos B + cos C is less than or equal to 2

    • one year ago
  19. mathslover Group Title
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    I got some conceptual answers previous time but I am searching for a numerical one.. @hartnn

    • one year ago
  20. ParthKohli Group Title
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    And mine was numerical but only for a right triangle =/

    • one year ago
  21. mathslover Group Title
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    =/ yeah @hartnn any try?

    • one year ago
  22. hartnn Group Title
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    if i get something, i'll post.

    • one year ago
  23. mathslover Group Title
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    \[{cos A + cos B + cos C = \\ \frac{a(b^2+c^2) + b(a^2+c^2) + \\c(a^2+b^2) - (a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{2abc}}\]

    • one year ago
  24. AravindG Group Title
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    trying on paper

    • one year ago
  25. mathslover Group Title
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    Thanks! I am also trying my best...

    • one year ago
  26. AravindG Group Title
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    got it !

    • one year ago
  27. mathslover Group Title
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    ?

    • one year ago
  28. AravindG Group Title
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    |dw:1356616135337:dw|

    • one year ago
  29. mathslover Group Title
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    yes..

    • one year ago
  30. AravindG Group Title
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    now cos A+cos B+Cos C is always less than equal to 3

    • one year ago
  31. mathslover Group Title
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    We have to prove for less than or equal to 2

    • one year ago
  32. AravindG Group Title
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    ya i am proving

    • one year ago
  33. mathslover Group Title
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    ok

    • one year ago
  34. AravindG Group Title
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    |dw:1356616236014:dw|

    • one year ago
  35. AravindG Group Title
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    you get that >?

    • one year ago
  36. mathslover Group Title
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    No :(

    • one year ago
  37. AravindG Group Title
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    why not ?where u have doubt ?

    • one year ago
  38. mathslover Group Title
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    cos A + cos B - cos(A+B) , how is it less than or equal to 3 ?

    • one year ago
  39. AravindG Group Title
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    i wrote cos A+cos B+cos(18-(A+B)<=3

    • one year ago
  40. AravindG Group Title
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    this is because maximum value of cos is 1

    • one year ago
  41. AravindG Group Title
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    ie 1+1+1=3

    • one year ago
  42. AravindG Group Title
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    all other cases cos a +cos b +cos c value will be less than 3

    • one year ago
  43. AravindG Group Title
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    and note i wrote c as 180-(A+B)

    • one year ago
  44. AravindG Group Title
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    @mathslover you get it?

    • one year ago
  45. mathslover Group Title
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    wait!

    • one year ago
  46. AravindG Group Title
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    ok take your time !

    • one year ago
  47. mathslover Group Title
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    Are you taking A = B = 0 degrees?

    • one year ago
  48. mathslover Group Title
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    since cos A + cos B + cos (180-(A+B)) = 3 is not possible ?

    • one year ago
  49. mathslover Group Title
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    I may be wrong but please correct me if I am ...

    • one year ago
  50. AravindG Group Title
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    yep for maximum value we take A=B=C=0 degree

    • one year ago
  51. mathslover Group Title
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    but A , B and C are the angles of a triangle..

    • one year ago
  52. AravindG Group Title
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    yess that was in general case

    • one year ago
  53. AravindG Group Title
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    in the triangle case angle c is replaced by 180-(A+B)

    • one year ago
  54. AravindG Group Title
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    hence cos C becomes cos (180-(A+B))

    • one year ago
  55. mathslover Group Title
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    Yes but we can not take the angles of a triangle as 0 degrees..

    • one year ago
  56. AravindG Group Title
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    i didnt told you to take !!i said that is general case!

    • one year ago
  57. AravindG Group Title
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    but equal to not possible

    • one year ago
  58. AravindG Group Title
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    in triangle you have equation cos A+cos B+cos(180-(A+B))<3

    • one year ago
  59. mathslover Group Title
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    so can cos A + cos B +cos C be greater than 2 also ?

    • one year ago
  60. AravindG Group Title
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    no ! thats where we apply trigonometry ! os(180-x)=-sin x

    • one year ago
  61. AravindG Group Title
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    you understand that much?

    • one year ago
  62. mathslover Group Title
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    cos ( 180 - x ) = - cos (x)

    • one year ago
  63. AravindG Group Title
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    oops srry ya

    • one year ago
  64. AravindG Group Title
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    -cos x

    • one year ago
  65. AravindG Group Title
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    so our equation for triangle becomes cos A+cos B-cos(A+B)<3

    • one year ago
  66. mathslover Group Title
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    got it now thanks a lot

    • one year ago
  67. AravindG Group Title
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    :) you are welcome bro!

    • one year ago
  68. hartnn Group Title
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    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

    • one year ago
  69. hartnn Group Title
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    ?

    • one year ago
  70. AravindG Group Title
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    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

    • one year ago
  71. hartnn Group Title
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    cos can be negative

    • one year ago
  72. hartnn Group Title
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    ohh..in 1st and 4th Quadrant cos is positive....

    • one year ago
  73. AravindG Group Title
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    ya whatever value cos takes it wont exceed +2

    • one year ago
  74. AravindG Group Title
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    yep !

    • one year ago
  75. hartnn Group Title
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    i thought cos(A+B) could be negative, thats why i doubted..but cos (A+B) will be positive...

    • one year ago
  76. AravindG Group Title
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    ;)

    • one year ago
  77. AravindG Group Title
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    its all about the quadrants :)

    • one year ago
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