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mathslover

  • 2 years ago

cos A + cos B + cos C where A B and C are the angles of any triangle

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  1. hba
    • 2 years ago
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    Kehna kia chate ho ?

  2. hartnn
    • 2 years ago
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    do we have options/choices for this ?

  3. mathslover
    • 2 years ago
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    iwant to simplify it...

  4. hba
    • 2 years ago
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    \[\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }\]

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

  6. hartnn
    • 2 years ago
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    cos A + cos B + cos C is in the most simplified form

  7. hba
    • 2 years ago
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    Yeah right :)

  8. hartnn
    • 2 years ago
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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.

  9. mathslover
    • 2 years ago
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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

  10. mathslover
    • 2 years ago
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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 : \[\large{\frac{-4s^3}{abc}}\]

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

  12. mathslover
    • 2 years ago
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    let me show you my complete work wait..

  13. hartnn
    • 2 years ago
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    that doesn't hold for an equilateral triangle, all sides equal, say =1

  14. mathslover
    • 2 years ago
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    Yes but I am getting this answer by simplifying cos A + cos B + cos C by using cosines law..

  15. hartnn
    • 2 years ago
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    ok, show your work, we'll find the error...

  16. mathslover
    • 2 years ago
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    ok 1 minute

  17. mathslover
    • 2 years ago
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    Oh! I think I got my mistake ... ..

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

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

  20. ParthKohli
    • 2 years ago
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    And mine was numerical but only for a right triangle =/

  21. mathslover
    • 2 years ago
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    =/ yeah @hartnn any try?

  22. hartnn
    • 2 years ago
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    if i get something, i'll post.

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

  24. AravindG
    • 2 years ago
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    trying on paper

  25. mathslover
    • 2 years ago
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    Thanks! I am also trying my best...

  26. AravindG
    • 2 years ago
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    got it !

  27. mathslover
    • 2 years ago
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    ?

  28. AravindG
    • 2 years ago
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    |dw:1356616135337:dw|

  29. mathslover
    • 2 years ago
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    yes..

  30. AravindG
    • 2 years ago
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    now cos A+cos B+Cos C is always less than equal to 3

  31. mathslover
    • 2 years ago
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    We have to prove for less than or equal to 2

  32. AravindG
    • 2 years ago
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    ya i am proving

  33. mathslover
    • 2 years ago
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    ok

  34. AravindG
    • 2 years ago
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    |dw:1356616236014:dw|

  35. AravindG
    • 2 years ago
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    you get that >?

  36. mathslover
    • 2 years ago
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    No :(

  37. AravindG
    • 2 years ago
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    why not ?where u have doubt ?

  38. mathslover
    • 2 years ago
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    cos A + cos B - cos(A+B) , how is it less than or equal to 3 ?

  39. AravindG
    • 2 years ago
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    i wrote cos A+cos B+cos(18-(A+B)<=3

  40. AravindG
    • 2 years ago
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    this is because maximum value of cos is 1

  41. AravindG
    • 2 years ago
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    ie 1+1+1=3

  42. AravindG
    • 2 years ago
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    all other cases cos a +cos b +cos c value will be less than 3

  43. AravindG
    • 2 years ago
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    and note i wrote c as 180-(A+B)

  44. AravindG
    • 2 years ago
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    @mathslover you get it?

  45. mathslover
    • 2 years ago
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    wait!

  46. AravindG
    • 2 years ago
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    ok take your time !

  47. mathslover
    • 2 years ago
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    Are you taking A = B = 0 degrees?

  48. mathslover
    • 2 years ago
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    since cos A + cos B + cos (180-(A+B)) = 3 is not possible ?

  49. mathslover
    • 2 years ago
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    I may be wrong but please correct me if I am ...

  50. AravindG
    • 2 years ago
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    yep for maximum value we take A=B=C=0 degree

  51. mathslover
    • 2 years ago
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    but A , B and C are the angles of a triangle..

  52. AravindG
    • 2 years ago
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    yess that was in general case

  53. AravindG
    • 2 years ago
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    in the triangle case angle c is replaced by 180-(A+B)

  54. AravindG
    • 2 years ago
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    hence cos C becomes cos (180-(A+B))

  55. mathslover
    • 2 years ago
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    Yes but we can not take the angles of a triangle as 0 degrees..

  56. AravindG
    • 2 years ago
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    i didnt told you to take !!i said that is general case!

  57. AravindG
    • 2 years ago
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    but equal to not possible

  58. AravindG
    • 2 years ago
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    in triangle you have equation cos A+cos B+cos(180-(A+B))<3

  59. mathslover
    • 2 years ago
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    so can cos A + cos B +cos C be greater than 2 also ?

  60. AravindG
    • 2 years ago
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    no ! thats where we apply trigonometry ! os(180-x)=-sin x

  61. AravindG
    • 2 years ago
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    you understand that much?

  62. mathslover
    • 2 years ago
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    cos ( 180 - x ) = - cos (x)

  63. AravindG
    • 2 years ago
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    oops srry ya

  64. AravindG
    • 2 years ago
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    -cos x

  65. AravindG
    • 2 years ago
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    so our equation for triangle becomes cos A+cos B-cos(A+B)<3

  66. mathslover
    • 2 years ago
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    got it now thanks a lot

  67. AravindG
    • 2 years ago
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    :) you are welcome bro!

  68. hartnn
    • 2 years ago
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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

  69. hartnn
    • 2 years ago
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    ?

  70. AravindG
    • 2 years ago
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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

  71. hartnn
    • 2 years ago
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    cos can be negative

  72. hartnn
    • 2 years ago
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    ohh..in 1st and 4th Quadrant cos is positive....

  73. AravindG
    • 2 years ago
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    ya whatever value cos takes it wont exceed +2

  74. AravindG
    • 2 years ago
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    yep !

  75. hartnn
    • 2 years ago
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    i thought cos(A+B) could be negative, thats why i doubted..but cos (A+B) will be positive...

  76. AravindG
    • 2 years ago
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    ;)

  77. AravindG
    • 2 years ago
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    its all about the quadrants :)

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