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

[╔(σ_σ)╝]

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\theta i sin\theta j
and
Q = cos\phi i sin\phi j
are unit vectors in the xy-plane, respectively, making angles \theta and \phi with the x-axis.

This was easy I already did this.

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

My solution:
I found that
P dot Q = cos(\phi - \theta)But the answer at the back of my book says
cos(\phi)cos( \theta) - sin(\phi)sin( \theta)
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(\phi + \theta) so I don't even know what they are trying to say.

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

My solution

|P - Q| = (P -Q) dot (P - Q) = P² +Q² -2(P dot Q)
= 2 - 2|P||Q|cos\varphi
= 2 - 2cos\varphi

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

Right ?

But then my book says the answer is |sin (0.5(\phi - \theta))|
So does anyone know what I'm doing incorrect ?

 

[tiny-tim]

Hi ╔(σ_σ)╝!

(have a theta: θ and a phi: φ )

… I found that
P dot Q = cos(\phi - \theta)


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

If that's meant to be cos(θ - φ), then the book is wrong.

3) … So |P - Q|/2 = 1-cos\varphi

Right ?

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

i] |A| is the square-root of A.A

ii] 1 + cos2θ = 2sin²θ

 

[╔(σ_σ)╝]

Hi ╔(σ_σ)╝!

(have a theta: θ and a phi: φ )If that's meant to be cos(θ - φ), then the book is wrong.

That is meant to be
cos(θ-\phi)

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

i] |A| is the square-root of A.A

ii] 1 + cos2θ = 2sin²θ

I see.

I was careless. I'll try again and see what I get.

 

[╔(σ_σ)╝]

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+ \phi) ?

 

[tiny-tim]

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

No, √(1 - cos(θ - φ))/2 is correct.

What about the maniplulation of P and Q to generate cos ( θ+ \phi) ?

How did you reformulate P and Q for that?

 

[╔(σ_σ)╝]

No, √(1 - cos(θ - φ))/2 is correct.


How did you reformulate P and Q for that?

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(\phi + \theta) so I don't even know what they are trying to say.

As I said previously.

Is there an easier way ?

 

[╔(σ_σ)╝]

Does anyone know how I can maniplulate P and Q to get cos(\phi +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)

 

[tiny-tim]

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

why minus?

(what basic formula are you using ??)

 

[╔(σ_σ)╝]

why minus?

(what basic formula are you using ??)

cos ( θ+\phi) = cos (θ)cos(\phi) - sin (θ)sin(\phi)

Right ?

Hence, the minus.

 

[tiny-tim]

cos ( θ+\phi) = cos (θ)cos(\phi) - sin (θ)sin(\phi)

Right ?

Hence, the minus.

That's the answer.

Why the minus in …

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

 

[╔(σ_σ)╝]

That's the answer.

Why the minus in …

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

Q dot e₁ I get cos\phiP dot e₂ I get sinθ

Q dot e₂ I get sin\phi

Combining all of these

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

(cosθ)(cos\phi) - (sinθ)(sin\phi)
Sorry I don't seem to see the problem.