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Proof of sin(θ-Φ)=sinθcosΦ-cosθsinΦ using vector algebra

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data given two unit vectors a= cosθi + sinθi b=cosΦi+sinΦj prove that sin(θ-Φ)=sinθcosΦ-cosΦsinθ using vector algebra


    2. Relevant equations sin(θ-Φ)=sinθcosΦ-cosΦsinθ


    3. The attempt at a solution axb= (cosθsinΦ-cosΦsinθ)k and I'm guessing that the change in sign has something to do with the fact that k is perpendicular to the vectors I'm usingΦ-θ
     
  2. jcsd
  3. Nov 7, 2015 #2

    haruspex

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    You could try a suitable rotation of axes, so that the cross product should still look the same.
     
  4. Nov 7, 2015 #3

    ehild

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    The vectors a and b have zero "z" component. Remember how the components of the cross product are calculated.
     
  5. Nov 8, 2015 #4
    when I calculated the components I got axb = ( cosθi +sinθj )x(cosΦi + sinΦj) axb=cosθcosΦixi + cosθsinΦixj +sinθcosΦjxi +sinθsinΦjxj ixi=1x1xsino=0 jxjxsin0=0 ixj=1x1sin90=1 and jxi=-1 because AxB=-BxA and was then left with axb= cosθsinΦ(1) + sinθcosΦ(-1)=cosθsinΦ -sinθcosΦ but its supposed to be the other way around, I dont understand where I'm going wrong :(
     
  6. Nov 8, 2015 #5

    ehild

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    Think of the other definition of the cross product, how to get its direction applying right-had rule, so sin(θ-φ)=bxa.
    https://www.mathsisfun.com/algebra/vectors-cross-product.html
     
  7. Nov 8, 2015 #6
    oh I think I get it now so what your saying is although I could find axb and get the answer I got first , I could just as easily choose bxa as they're both the same but have opposite signs and so it still satisfies the proof.
     
  8. Nov 8, 2015 #7

    ehild

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    The correct one is bxa.
     
  9. Nov 8, 2015 #8
    Isn't (i + j) X (i + j) = j X i + i X j)
    What is the sign of j X i ?
     
  10. Nov 8, 2015 #9

    ehild

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    minus. Why do you ask? It was correct in Post#4.
     
    Last edited: Nov 8, 2015
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