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$$\phi$$P 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.