Dot product and its representation as sum. 
If I define the dot product $u\cdot v $ where $u,v\in \mathbb{R}^n$ as $u\cdot v= |u||v|\cos \theta$ where $\theta$ is the angle between $u$ and $v$. How can I get that 
  $$u\cdot v= u_1v_1+u_2v_2+u_3v_3 $$
  with $u=u_1e_1+u_2e_2+\cdots+u_ne_n$ and $v=v_1e_1+v_2e_2+\cdots+v_ne_n$? 

$e_1... e_n$ is the canonical base. 
If we think in $\mathbb{R}^2$ we can suppose that, $u$ have an angle of $\alpha$  with the X axis, $v$ have an angle $\beta$ with the $X$ axis, and $\alpha\geq\beta$. Then $\theta=\alpha-\beta$. We get
$$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta $$
Using the relation $cos\alpha= \displaystyle\frac{u_1}{|u|}, \sin \alpha=\displaystyle\frac{u_2}{|u|}$ and analogous  relation for $v$
$$\cos\theta= \displaystyle\frac{u_1}{|u|}\displaystyle\frac{v_1}{|v|}+\displaystyle\frac{u_2}{|u|}\displaystyle\frac{v_2}{|v|}$$
multiplying by |u||v| we get the result. 
Can we generalize this idea to $\mathbb{R}^n$?
Thanks!

Letting $\vec u = (u_1, u_2, \dots, u_n)$ and 
        $\vec v = (v_1, v_2, \dots, v_n)$,
then $\displaystyle |u|^2 = \sum_{i=1}^n u_i^2$,
     $\displaystyle |v|^2 = \sum_{i=1}^n v_i^2$, and
     $\displaystyle |u-v|^2 = \sum_{i=1}^n (u_i - v_i)^2$.
By the law of cosines, 
$\cos \theta = \dfrac{|u|^2 + |v|^2 - |u-v|^2}{2\,|u|\,|v|}$. So
\begin{align}
   u \circ v
   &= |u|\,|v|\,\cos \theta \\
   &= \dfrac 12(|u|^2 + |v|^2 - |u-v|^2) \\
   &= \dfrac 12\left( 
     \sum_{i=1}^n u_i^2 + 
     \sum_{i=1}^n v_i^2 - 
     \sum_{i=1}^n (u_i^2 + v_i^2 - 2u_iv_i)\right) \\
   &= \sum_{i=1}^n u_iv_i
\end{align}
