From $\|\vec u\|\|\vec v\|\cos(\theta)$ to the dot product definition I know that $\vec u\cdot \vec v\overset{\text{def}}{=}u_1v_1+u_2v_2+\cdots+u_nv_n$ and I know the proof of $\vec u\cdot \vec v=\|\vec u\|\|\vec v\|\cos(\theta)$. $$$$my question is, if I'm redefining the dot product to be $\vec u\cdot \vec v\overset{\text{def}}{=}\|\vec u\|\|\vec v\|\cos(\theta)$ can I proof that $\vec u\cdot \vec v=u_1v_1+u_2v_2+\cdots+u_nv_n$ is true?
Edit
Some people said I need to redefine $\|\vec u\|$, so I'm defining it with pythagorean theorem so it's value won't change
 A: By Cauchy Schwartz there exists a unique $\theta$ such that for a given inner product, 
$$ \langle u,v \rangle  = \| u \| \|v\| \cos(\theta)$$
is well defined where $\|u\| = \langle u,u \rangle$. The dot product is a particular inner product so yes the result follows immediately if you know these facts. 
A: \begin{align} \text{using Cosine Law}& \\  \|\vec u-\vec v\|^2=&\|\vec u\|^2+\|\vec v\|^2-2\|\vec u\|\|\vec v\|\cos\theta\\ \|\vec u\|\|\vec v\|\cos\theta=&\frac{1}{2}\left(\|\vec u\|^2+\|\vec v\|^2-\|\vec u-\vec v\|^2\right)\\\text{so: }&\\ \vec u \cdot \vec v =& \|\vec u\|\|\vec v\|\cos \theta \\ =& \frac{1}{2}\left(\|\vec u\|^2 + \|\vec v\|^2 - \|\vec u-\vec v\|^2\right) \\ &\text{i defined $\|\vec u\|$ with pythagorean theorem so:}\\=&\frac{1}{2}\left( \sum_{i=1}^n u_i^2 + \sum_{i=1}^n v_i^2 - \sum_{i=1}^n (u_i-v_i)^2\right) \\ =&\frac{1}{2}\left( \sum_{i=1}^n u_i^2 + \sum_{i=1}^n v_i^2 - \sum_{i=1}^n \left(u_i^2+v_i^2-2u_iv_i\right)\right)\\ =&\frac{1}{2}\left( \sum_{i=1}^n\left( u_i^2 + v_i^2 - \left(u_i^2+v_i^2-2u_iv_i\right)\right)\right)\\ =& \sum_{i=1}^n u_iv_i \end{align}
thanks to the comments that helped
