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

• ╔(σ_σ)╝
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.
╔(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 ?

Hi ╔(σ_σ)╝!

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

tiny-tim said:
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.

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+ $$\phi$$) ?

╔(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+ $$\phi$$) ?

How did you reformulate P and Q for that?

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

As I said previously.

Is there an easier way ?

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)

╔(σ_σ)╝ said:
(P dot e1)(Q dot e1) - (P dot e2)(Q dot e2)

why minus?

(what basic formula are you using ??)

tiny-tim said:
why minus?

(what basic formula are you using ??)
cos ( θ+$$\phi$$) = cos (θ)cos($$\phi$$) - sin (θ)sin($$\phi$$)

Right ?

Hence, the minus.

╔(σ_σ)╝ said:
cos ( θ+$$\phi$$) = cos (θ)cos($$\phi$$) - sin (θ)sin($$\phi$$)

Right ?

Hence, the minus.

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

tiny-tim said:

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.

## 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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