calculus - Derive the formula for the cosine of the difference of two angles from the dot product formula

My book states that:

The formula for the cosine of the difference of two angles is deduced
as an application of the scalar product of two vectors:

$$\cos(\alpha - \beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta$$

From this formula, we can deduce the formula for the sine of the
difference:

$$\sin(\alpha-\beta)=\cos[\frac{\pi}{2}-(\alpha-\beta)]=\\\cos[\frac{\pi}{2}-\alpha-(-\beta)] =\\ \cos(\frac{\pi}{2}-\alpha)\cos(-\beta)+\sin(\frac{\pi}{2}-\alpha)\sin(-\beta)$$

$$\\\\$$

$$\\\\\\\\\\\\\sin(\alpha-\beta )= \sin\alpha\cos\beta-\cos\alpha\sin\beta$$

Deduce the following expression starting from the formulas above:

• $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$

I have two questions:

1. I don't understand how my book means by

The formula for the cosine of the difference of two angles is deduced as an application of the scalar product of two vectors:

Could you explain to me what this means?

2. How do I solve the given problem?

Answer

Draw two position vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$ with unit magnitude and at angles $\alpha, \beta$ to the positive $x$-axis. Then the angle between the two is $\alpha - \beta$ (assuming $\alpha > \beta$ w.l.o.g). But $\mathbf{v}_1 \cdot \mathbf{v}_2$ is the cosine of the angle between them. So $\cos (\alpha - \beta) = \mathbf{v}_1 \cdot \mathbf{v}_2$.

But remember that the two vectors lie on the unit circle and have components $\mathbf{v}_1 = \cos \alpha \mathbf{i} + \sin \alpha \mathbf{j}$ and $\mathbf{v}_2 = \cos \beta \mathbf{i} + \sin \beta \mathbf{j}$. By the definition of the dot product

$$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta.$$

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To solve the given problem: note that $\beta \mapsto -\beta$ gives

$$\cos (\alpha - (-\beta)) = \cos \alpha \cos (-\beta) + \sin \alpha \sin (-\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta)$$ using the oddity of $\sin$ and even-ness of $\cos$.