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

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In summary: What I meant was, why did you choose to use a minus instead of a plus in your manipulation? Also, in the final answer (cosθ)(cosφ) - (sinθ)(sinφ), the minus sign is actually a plus sign because of the minus in front of (P dot e2)(Q dot e2). So the final answer is actually (cosθ)(cosφ) + (sinθ)(sinφ). I hope that clears things up.
  • #1
╔(σ_σ)╝
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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 ?
 
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  • #2
Hi ╔(σ_σ)╝! :smile:

(have a theta: θ and a phi: φ :wink:)
╔(σ_σ)╝ said:
… 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.

If that's meant to be cos(θ - φ), then the book is wrong. :frown:
3) … 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]))|

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

ii] 1 + cos2θ = 2sin2θ :wink:
 
  • #3


tiny-tim said:
Hi ╔(σ_σ)╝! :smile:

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

That is meant to be
cos(θ-[tex]\phi[/tex])

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 :wink:
I see.

I was careless. I'll try again and see what I get.
 
  • #4


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+ [tex] \phi[/tex]) ?
 
  • #5
╔(σ_σ)╝ said:
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+ [tex] \phi[/tex]) ?

How did you reformulate P and Q for that?
 
  • #6


tiny-tim said:
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([tex]\phi[/tex] + [tex]\theta[/tex]) so I don't even know what they are trying to say.

As I said previously.

Is there an easier way ?
 
  • #7


Does anyone know how I can maniplulate P and Q to get cos([tex]\phi[/tex] +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
╔(σ_σ)╝ said:
(P dot e1)(Q dot e1) - (P dot e2)(Q dot e2)

why minus? :confused:

(what basic formula are you using ??)
 
  • #9


tiny-tim said:
why minus? :confused:

(what basic formula are you using ??)
cos ( θ+[tex]\phi[/tex]) = cos (θ)cos([tex]\phi[/tex]) - sin (θ)sin([tex]\phi[/tex])

Right ?

Hence, the minus.
 
  • #10
╔(σ_σ)╝ said:
cos ( θ+[tex]\phi[/tex]) = cos (θ)cos([tex]\phi[/tex]) - sin (θ)sin([tex]\phi[/tex])

Right ?

Hence, the minus.

That's the answer. :confused:

Why the minus in …
╔(σ_σ)╝ said:
(P dot e1)(Q dot e1) - (P dot e2)(Q dot e2)
 
  • #11


tiny-tim said:
That's the answer. :confused:

Why the minus in …

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

Q dot e1 I get cos[tex]\phi[/tex]P dot e2 I get sinθ

Q dot e2 I get sin[tex]\phi[/tex]

Combining all of these

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

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

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

1. What is the formula for cos( theta 1 + theta2)?

The formula for cos( theta 1 + theta2) is cos( theta 1) * cos( theta2) - sin( theta 1) * sin( theta2).

2. How do you obtain the formula for cos( theta 1 + theta2)?

To obtain the formula for cos( theta 1 + theta2), we use the trigonometric identity cos(a + b) = cos(a) * cos(b) - sin(a) * sin(b) and substitute theta 1 and theta2 for a and b.

3. What is the significance of cos( theta 1 + theta2) in trigonometry?

Cos( theta 1 + theta2) is used to find the cosine value of the sum of two angles, which can be useful in solving problems involving triangles and other shapes in trigonometry.

4. Can the formula for cos( theta 1 + theta2) be simplified?

Yes, the formula for cos( theta 1 + theta2) can be simplified using other trigonometric identities, such as the double angle formula cos(2a) = cos^2(a) - sin^2(a).

5. How can the formula for cos( theta 1 + theta2) be applied in real-life situations?

The formula for cos( theta 1 + theta2) can be applied in fields such as engineering, physics, and astronomy to calculate angles and distances in various scenarios. It can also be used in navigation, surveying, and mapping.

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