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:


*

*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?


*How do I solve the given problem?


If you have two vectors $v=(v_1,v_2)$ and $w=(w_1,w_2)$, then their dot product is $$v \cdot w = v_1 w_1 + v_2 w_2.$$ Additionally, we also have
$$v \cdot w = \|v\| \|w\| \cos \theta$$ where $\|v\|=\sqrt{v_1^2+v_2^2}$, $\|w\|=\sqrt{w_1^2+w_2^2}$, and $\theta$ is the angle between $v$ and $w$. Proof here.
Now, consider $v=(\cos \alpha, \sin \alpha)$ and $w=(\cos \beta, \sin \beta)$. Do you now see why the formula for $\cos(\alpha-\beta)$ then follows? Hint: we have $\|v\|=\|w\|=1$ and $\theta=\alpha-\beta$.

For your second question,note $\cos(\alpha+\beta)=\cos(\alpha-(-\beta))$,
then apply your formula for the cosine of a difference of angles, and then use the fact that $\sin$ is an odd function, and $\cos$ is an even function. This has been done for you in the comments.
