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Proving Trigonometry equation

  1. Oct 2, 2011 #1
    I'm new to this forum, anyhow thanks for the admin making such forum, doesn't matter !

    I've been trying this calculation for one day still no answer,

    1. The problem statement, all variables and given/known data

    cos (A/2)+cos (B/2)+cos (C/2)=4cos [(B+C)/4]*cos [(C+A)/4]*cos [(A+B)/4]

    A+B+C = pi

    2. Relevant equations




    3. The attempt at a solution

    Cos C+Cos D=2cos[(C+D)/2]*cos[(C-D)/2]

    with above equation tried no use !
     
    Last edited: Oct 2, 2011
  2. jcsd
  3. Oct 2, 2011 #2

    Filip Larsen

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  4. Oct 2, 2011 #3
  5. Oct 2, 2011 #4
    no one to help me?
     
  6. Oct 2, 2011 #5

    Mark44

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    What is D?

    Although you didn't say this, it looks like you are supposed to show that the equation above is an identity. So there is no "calculation" to do - just show that the left and right sides of the equation are the same for any values of A, B, and C that add up to [itex]\pi[/itex] radians.

    Since the assumption is that A + B + C = [itex]\pi[/itex], you can replace A + B on the right side with [itex]\pi[/itex] - C, and do something similar with C + A and with B + C.
     
  7. Oct 3, 2011 #6
    D is nothing it just a formula to solve the above ! btw thanks for ur help, I got a simple clue
    let me try !
     
  8. Oct 3, 2011 #7

    NascentOxygen

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    Ah! I missed that line with A+B+C=Pi
    That changes things!
     
  9. Oct 3, 2011 #8

    SammyS

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    There may be a simpler method. The way I did it is somewhat involved.

    cos x + cos y = 2cos[(x+y)/2]*cos[(x-y)/2] is a place to start.

    Then, cos (A/2)+cos (B/2) = 2cos[(A/2+B/2)/2]*cos[(A/2-B/2)/2] = 2cos(A/4+B/4)*cos(A/4-B/4)

    Also, notice that A/2 + B/2 + C/2 = pi/2 , so that C/2 = π/2 - (A/2 + B/2) .

    Therefore, cos (C/2) = cos(π/2 - (A/2 + B/2)) = sin(A/2 + B/2) = sin(2(A/4 + B/4)) .

    And sin(2(A/4 + B/4)) = 2sin(A/4 + B/4)*cos(A/4 + B/4) .

    Now you have cos (A/2)+cos (B/2)+cos (C/2) = 2cos(A/4+B/4)*cos(A/4-B/4) + 2sin(A/4 + B/4)*cos(A/4 + B/4) .

    Factor out 2cos(A/4+B/4) .

    That's a start. Still a ways to go.
     
  10. Oct 4, 2011 #9

    ehild

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    I would do it just backwards... The proof is very simple if you apply the identity cos(x)cos(y)=cos(x+y)+cos(x-y) to the right-hand side twice.

    ehild
     
    Last edited: Oct 4, 2011
  11. Oct 5, 2011 #10

    ehild

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    I go ahead for those who read this thread. I will rename the variables, so A/2=a, B/2=b, C/2=c.

    Prove that

    cos(a)+cos(b)+cos(c)=4cos(0.5(a+b))*cos(0.5(a+c))* cos(0.5 (c+b)), if a+b+c=pi/2.


    Using the identity cosxcosy=(cos(x+y)+cos(x-y))/2

    cos(0.5(a+c))* cos(0.5 (b+c))=0.5[cos( 0.5(a+b+2c))+cos(0.5(a-b))],

    multiplying with 4cos(0.5(a+b)):

    2cos(0.5(a+b))*[ cos(0.5(a+b+2c))+cos(0.5(a-b))]=

    =[cos(0.5(2a+2b+2c))+ cos(0.5(a+b+2c-(a+b))]+[ cos(0.5(a+b)+0.5(a-b))+ cos(0.5(a+b)-0.5(a-b))]=

    =cos(a+b+c)+cos(c)+cos(a)+cos(b)=cos(pi/2)+cos(a)+cos(b)+cos(c) ,

    equal to the LHS of the original equation.



    ehild
     
    Last edited: Oct 5, 2011
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