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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 ?
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 ?