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Homework Help: 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


    Staff: Mentor

    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


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    Staff: Mentor

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


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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


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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.

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


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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.

    Last edited: Oct 5, 2011
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