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

1. Dec 26, 2009

### ╔(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 ?

2. Dec 26, 2009

### tiny-tim

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

ii] 1 + cos2θ = 2sin2θ

3. Dec 26, 2009

### ╔(σ_σ)╝

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

That is meant to be
cos(θ-$$\phi$$)

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.

4. Dec 26, 2009

### ╔(σ_σ)╝

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+ $$\phi$$) ?

5. Dec 26, 2009

### tiny-tim

No, √(1 - cos(θ - φ))/2 is correct.
How did you reformulate P and Q for that?

6. Dec 26, 2009

### ╔(σ_σ)╝

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

As I said previously.

Is there an easier way ?

7. Dec 27, 2009

### ╔(σ_σ)╝

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

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)

8. Dec 27, 2009

### tiny-tim

why minus?

(what basic formula are you using ??)

9. Dec 27, 2009

### ╔(σ_σ)╝

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

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

Right ?

Hence, the minus.

10. Dec 27, 2009

### tiny-tim

Why the minus in …

11. Dec 27, 2009

### ╔(σ_σ)╝

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$$\phi$$

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.