Derive cos(A-B) from the law of cosines

[Karol]

Homework Statement

In the chapter about differentiation there is this question:

Homework Equations

Cosine rule:
$$a^2=b^2+c^2-2bc\cos\alpha$$
$$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$

The Attempt at a Solution

$$a^2=b^2+c^2-2bc\cos\alpha$$
$$d^2=e^2+c^2-2ec\cos\beta$$
$$(a-d)^2=b^2+e^2-2bc\cos(\alpha-\beta)$$
It doesn't lead to the solution

 

[Mark44]

Homework Statement

In the chapter about differentiation there is this question:
View attachment 206543

Homework Equations

Cosine rule:
$$a^2=b^2+c^2-2bc\cos\alpha$$
$$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$

The Attempt at a Solution

View attachment 206553
$$a^2=b^2+c^2-2bc\cos\alpha$$
$$d^2=e^2+c^2-2ec\cos\beta$$
$$(a-d)^2=b^2+e^2-2bc\cos(\alpha-\beta)$$
It doesn't lead to the solution

I wouldn't set it up the way you have in your drawing. Sketch the unit circle with one ray making an angle of $\alpha$ going out to the point $(\cos \alpha, \sin \alpha)$ on the unit circle. Draw another ray with a smaller angle of $\beta$ going out to the point $(\cos \beta, \sin \beta)$ on the unit circle.

Use the Law of Cosines to find the distance between those two points on the unit circle. The result for $\cos(\alpha - \beta)$ is pretty easy after that.

 

[Karol]

$$a^2=2(1-\cos\alpha),~~b^2=2(1-\cos\beta)$$
$$x^2=2(1-\cos(\alpha-\beta))$$
$$x\neq a-b$$
Even if x were (a-b) it's complicated

 

[willem2]

View attachment 206599
$$x^2=2(1-\cos(\alpha-\beta))$$
$$x\neq a-b$$
Even if x were (a-b) it's complicated

But you can compute what x is, because you know where the line segment x starts and ends. You only need pythagoras to do that.
Then you use the law of cosines to find Cos(A-B) from x.

 

[Karol]

$$x^2=(\sin\alpha-\sin\beta)^2+(\cos\beta-\cos\alpha)^2=2-2\cos(\alpha-\beta)$$
$$\Rightarrow\cos(\alpha-\beta)=\sin\alpha\cdot\sin\beta+\cos\alpha\cdot \cos\beta$$