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Stuck :s

  1. Oct 16, 2005 #1
    Hey
    So, I have to show that in a triangle, we have : (A, B, C are the angles of that triangle)

    sinB+sinC-sinA=4cos(A/2)sin(B/2)sin(C/2)

    okay, here is what I got to, then I got stuck
    i got it equal to:

    sinB+sinC-sinA = 2cos(A/2)cos(B/2)cos(C/2)+2cos(A/2)sin(B/2)sin(C/2)-2sin(B/2)sin(C/2)

    I don't know what to do next! Any hints?
     
    Last edited: Oct 16, 2005
  2. jcsd
  3. Oct 16, 2005 #2

    Fermat

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    You know that A + B + C = π

    or,

    A/2 = π/2 - (B+C)/2

    hence,

    sin(A/2) = sin(π/2 - (B+C)/2) = cos((B+C)/2)

    and,

    cos(A/2) = cos(π/2 - (B+C)/2) = sin((B+C)/2)

    I wouldn't use these identities in the expression you already have, but rather start from the beginning again, keeping these identies in mind.
     
  4. Oct 16, 2005 #3
    yes, that's what I did to get to the equation above. But I dont know what to do after...! So it can be equal to 4cos(A/2)sin(B/2)sin(C/2)
     
    Last edited: Oct 16, 2005
  5. Oct 16, 2005 #4

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    sinB + sinc - sinA =
    2sin((B+C)/2)cos((B-C)/2) - 2sin(A/2)cos(A/2)

    now use those identities
     
  6. Oct 16, 2005 #5
    yup, i used that before, and didn't get to anything good.
    Here is what I could do with it:

    sinB+sinC-sinA =
    2cos(A/2)cos(B/2)cos(C/2)+2cos(A/2)sin(B/2)sin(C/2)+2sin(A/2)cos(B/2)cos(C/2)-2sin(A/2)sin(B/2)sin(C/2)
     
  7. Oct 16, 2005 #6

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    sinB + sinc - sinA =
    2sin((B+C)/2)cos((B-C)/2) - 2sin(A/2)cos(A/2)

    using the identities, we get,

    2cos(A/2).cos((B-C)/2) - 2sin(A/2).cos(A/2)
    2cos(A/2){cos((B-C)/2) - sin(A/2)}

    using the identities again,

    2cos(A/2){cos((B-c)/2) - cos((B+C)/2)}

    now use the formula for the difference of two cosines,

    2cos(A/2){2sin(B/2)sin(C/2)}
    4cos(A/2)sin(B/2)sin(C/2)
    ===================
     
  8. Oct 16, 2005 #7
    OH YES! Sorry I didnt notice that I should use the formula for the difference of two cosines... Thanks a lot!
     
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