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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

    SammyS

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    What are the components of vector, a, in terms of a (the magnitude of the vector) and the angle α?

    What are the components of vector, b, in terms of b (the magnitude of the vector) and the angle β?
     
  4. Aug 26, 2012 #3
    The components of a are ax*cos(α) and ay*sin(α)

    The components of b are bx*cos(β) and by*sin(β)

    Thanks for reminding me about that I'll see where I can get from here
     
  5. Aug 26, 2012 #4

    SammyS

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    Not quite right.

    The components of a are ax = a*cos(α) and ay = a*sin(α)

    etc.

    B.T.W. ax = a1, etc. in your Relevant equations for the dot & cross products.
     
  6. Aug 26, 2012 #5

    vela

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    What are a1, a2, b1, and b2 in terms of a, b, α, and β?
     
  7. Aug 26, 2012 #6
    a1 = acos(α)
    a2 = asin(α)

    b1 = bcos(β)
    b2 = bsin(β)

    a dot b = acos(α)bcos(β) + asin(α)bsin(β)

    which pretty much completes the proof for a)
     
  8. Aug 26, 2012 #7

    SammyS

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    Do something similar for (b).

    ( a × b )3 = a1*b2 - a2*b1 . If both vectors are in the xy-plane, then the other components are zero.
     
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