(adsbygoogle = window.adsbygoogle || []).push({}); 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 e_{1})(Q dot e_{1}) - (P dot e_{2})(Q dot e_{2})

Where e_{1}= (1,0) and e_{2}= (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) = P^{2}+Q^{2}-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 ?

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

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