prove dot product in Rn equals lengths time cosine of angle Well, its known to everyone that $$x,y \in\Bbb{R}^n, <x, y>= \sum{x_iy_i}$$. It is how we define inner product in $\Bbb{R}^n$. But we also know that it has geometric interpretation like $$<x, y> = \Vert x\Vert\Vert y\Vert cos{\theta} $$. My question is how you prove this is true? This question seems quite simple but It's surprising I don't find an answer trying to fix this problem via google. 
It really bothers me sometimes because I've already gotten bachelor degree and I can't fix a straight forward problem like this....although im not in math major.
 A: Use the cosine rule.
Consider the triangle with vertices $0$, $x$ and $y$. It has side-lengths
$a=\sqrt{\langle x,x\rangle}$, $b=\sqrt{\langle y,y\rangle}$ and $c=\sqrt{\langle x-y,x-y\rangle}$. Then
$$\cos\theta=\frac{a^2+b^2-c^2}{2ab}.$$
Now put in the above expressions for $a$, $b$ and $c$.
A: For arbitrary $n$, this is in fact taken as definition for the angle between vectors.
Nevertheless, whatever we mean by angle between vectors, it should be invariant under isometries, which are by definition linear maps $A\colon\Bbb R^n\to \Bbb R^n$ that leave all lengths invariant, i.e., $\|Ax\|=\|x\|$ for all $x$. Then from $\langle x,y\rangle=\frac12(\| x+y\|^2-\|x\|^2-\|y\|^2)$, we see that isometries also respect scalar products, i.e., $\langle Ax,Ay\rangle=\langle x,y\rangle$ for all $x,y$.
One readily shows that there exists an isometry that takes any given vector $x\ne 0$ to a positive multiple of the first standard base vector $e_1$ and additionally takes $y$ to the span of $e_1$ and $e_2$. 
Hence we can restrict our investigation of angles to the $\Bbb R^2$ spanned by the first two base vectors and need only ask: How can we find the angle between vectors $x=(u,0)$ with $u>0$ and $y=(v,w)\ne(0,0)$? The latter vector can be written as $(v,w)=(r\cos\phi,r\sin\phi)$ with $r=\sqrt{v^2+w^2}$ and the desired $\phi$. Comparing with $\|x\|^2=u^2$, $\|y\|^2=v^2+w^2=r^2$, and $\langle x,y\rangle = uv$, we see that in this special case (and per isometries: always) $\langle x,y\rangle=\|x\|\|y\|\cos\phi$.
Note that the above gives us only $\phi\in[0,\pi]$, not in $[0,2\pi)$. Indeed, while we could distinguish angles in $(0,\pi)$ from those in $(\pi,2\pi)$ by the sign of $w$, and we could exploit this for the case $n=2$ by requiring orientation-preserving isometries. But this makes no sense as soon as we are in $n\ge 3$ dimensions: There, we will always have another orientation-reserving isometry that produces $(u,0)$ and $(v,-w)$.
