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Proving trig identities with dot and cross products

  1. Aug 26, 2012 #1
    1. The problem statement, all variables and given/known data

    The two vectors a and b lie in the xy plane and make angles α and β with the x axis.

    a)By evaluating ab in two ways (Namely ab = abcos(θ) and ab = a1b1+a2b2) prove the well-known trig identity
    cos(α-β)=cos(α)cos(β)+sin(α)sin(β)

    b)By similarly evaluating a X b prove that
    sin(α-β) = sin(α)cos(β) - cos(α)sin(β)

    c)Now let vector a make an angle -α with the x axis and find a similar expression for
    cos(α+β)


    2. Relevant equations

    ab = abcos(θ)

    ab = a1b1+a2b2



    3. The attempt at a solution

    I drew the vectors a and b with their appropriate angles to the x-axis... The angle between the vectors is (α-β) so I have ab = abcos(α-β) but I have no idea how to relate this to the trig identities!
     
  2. jcsd
  3. Aug 26, 2012 #2
    Hello ,
    Since you are stuck at a.b=|a||b|cos(α-β) ,
    Why dont you take use of the other equation you know a • b = a1b1+a2b2
    a1 , a2 are components along i and j . Does that ring any bell?
     
  4. Aug 26, 2012 #3
    Yes that makes sense but I'm not sure how to relate that to the trig identity :/
     
  5. Aug 26, 2012 #4
    can you write a1 ,a2 , b1 ,b2 in terms of |a| or |b| and some trigo function ?
     
  6. Aug 26, 2012 #5
    a1=ax*cos(α)

    a2=ay*sin(α)

    b1=bx*cos(β)

    b2=by*sin(β)

    I'm currently playing around with these but any headers would be appreciated!
     
  7. Aug 26, 2012 #6
    you have a vector in terms of i , j just do the dot product and equate it two what you already have from equation 1 .
    NOTE: you have posted this thread in multiple sections , moderators wont like it .
     
  8. Aug 26, 2012 #7
    Thanks.

    Sorry I didn't know where to put this...
     
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