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Obtain the forumal of cos( theta 1 + theta2) .

  1. Dec 26, 2009 #1
    Obtain the forumal of cos( theta 1 + theta2).....

    This should be a very basic question. I'm reviewing some linear algebra for my Field theory course but there is a question that I'm suck on.

    It says
    (a)
    Prove that
    P = cos[tex]\theta[/tex] i sin[tex]\theta[/tex] j
    and
    Q = cos[tex]\phi[/tex] i sin[tex]\phi[/tex] j
    are unit vectors in the xy-plane, respectively, making angles [tex]\theta[/tex] and [tex]\phi[/tex] with the x-axis.

    This was easy I already did this.

    (b)
    By means of dot product, obtain the formula for cos([tex]\phi[/tex] - [tex]\theta[/tex]). By similarly formulating P and Q, obtain the formula for cos([tex]\phi[/tex] + [tex]\theta[/tex])

    My solution:
    I found that
    P dot Q = cos([tex]\phi[/tex] - [tex]\theta[/tex])


    But the answer at the back of my book says
    cos([tex]\phi[/tex])cos( [tex]\theta[/tex]) - sin([tex]\phi[/tex])sin( [tex]\theta[/tex])
    I guess this is the same P dot Q, so my answer is the same as the book.


    But I can't seem to find a simple manipulation for P and Q to get the second formula all I can find is

    (P dot e1)(Q dot e1) - (P dot e2)(Q dot e2)

    Where e1 = (1,0) and e2 = (0,1)

    Again the back of my book gives the expanded for of cos([tex]\phi[/tex] + [tex]\theta[/tex]) so I don't even know what they are trying to say.




    3) The third part says if [tex]\varphi[/tex] is the angle P and Q, find |P -Q|/2 in terms of cos([tex]\phi[/tex] + [tex]\theta[/tex])

    My solution


    |P - Q| = (P -Q) dot (P - Q) = P2 +Q2 -2(P dot Q)
    = 2 - 2|P||Q|cos[tex]\varphi[/tex]
    = 2 - 2cos[tex]\varphi[/tex]

    So |P - Q|/2 = 1-cos[tex]\varphi[/tex]

    Right ?

    But then my book says the answer is |sin (0.5([tex]\phi[/tex] - [tex]\theta[/tex]))|



    So does anyone know what I'm doing incorrect ?
     
  2. jcsd
  3. Dec 26, 2009 #2

    tiny-tim

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    Hi ╔(σ_σ)╝! :smile:

    (have a theta: θ and a phi: φ :wink:)
    If that's meant to be cos(θ - φ), then the book is wrong. :frown:
    i] |A| is the square-root of A.A

    ii] 1 + cos2θ = 2sin2θ :wink:
     
  4. Dec 26, 2009 #3
    Re: Obtain the forumal of cos( theta 1 + theta2).....

    That is meant to be
    cos(θ-[tex]\phi[/tex])

    Btw I made a mistake it should be a + infront of the first sin .



    I see.

    I was careless. I'll try again and see what I get.
     
  5. Dec 26, 2009 #4
    Re: Obtain the forumal of cos( theta 1 + theta2).....

    Is there anything else I did incorrectly ? Apart from sqr(A dot A ) = |A| ?


    What about the maniplulation of P and Q to generate cos ( θ+ [tex] \phi[/tex]) ?
     
  6. Dec 26, 2009 #5

    tiny-tim

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    No, √(1 - cos(θ - φ))/2 is correct.
    How did you reformulate P and Q for that?
     
  7. Dec 26, 2009 #6
    Re: Obtain the forumal of cos( theta 1 + theta2).....

    As I said previously.

    Is there an easier way ?
     
  8. Dec 27, 2009 #7
    Re: Obtain the forumal of cos( theta 1 + theta2).....

    Does anyone know how I can maniplulate P and Q to get cos([tex]\phi[/tex] +θ).


    So far I was able to I did

    (P dot e1)(Q dot e1) - (P dot e2)(Q dot e2)

    Where e1 = (1,0) and e2 = (0,1)
     
  9. Dec 27, 2009 #8

    tiny-tim

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    why minus? :confused:

    (what basic formula are you using ??)
     
  10. Dec 27, 2009 #9
    Re: Obtain the forumal of cos( theta 1 + theta2).....


    cos ( θ+[tex]\phi[/tex]) = cos (θ)cos([tex]\phi[/tex]) - sin (θ)sin([tex]\phi[/tex])

    Right ?

    Hence, the minus.
     
  11. Dec 27, 2009 #10

    tiny-tim

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    That's the answer. :confused:

    Why the minus in …
     
  12. Dec 27, 2009 #11
    Re: Obtain the forumal of cos( theta 1 + theta2).....

    IF I take the dot product of P and e1 i get cosθ

    Q dot e1 I get cos[tex]\phi[/tex]


    P dot e2 I get sinθ

    Q dot e2 I get sin[tex]\phi[/tex]

    Combining all of these

    (P dot e1)(Q dot e1) - (P dot e2)(Q dot e2)

    (cosθ)(cos[tex]\phi[/tex]) - (sinθ)(sin[tex]\phi[/tex])



    Sorry I don't seem to see the problem.
     
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