Suppose $x_{\theta} = (\cos(\theta),\sin(\theta )) \in \mathbb R^2$. Prove $||x_{\theta + \theta^{'}} - x_{\theta}|| = ||x_{\theta^{'}} - x_{0}||$. 
Suppose $x_{\theta} = (\cos(\theta),\sin(\theta )) \in \mathbb R^2$. 
Prove $||x_{\theta + \theta^{'}} - x_{\theta}|| = ||x_{\theta^{'}} - x_{0}||$.

How do I prove that these values are equal corresponding to the distance between two pairs of points, with common angle, lying on the unit circle.
I want to prove that these numbers are equal: $||x_{\theta + \theta^{'}} - x_{\theta}||, ||x_{\theta^{'}} - x_{0}||$, so please don't give a geometric proof saying that similar triangles has the same side length.
This identity is used in a book to prove the identity $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ and later the similar identity for $\sin(a+b)$, so please don't use these in an answer.

 A: Here's a try. Let $x : [0,2\pi] \to \Bbb{R}^2$ be the arclength parametrization of the unit circle with $x(0)=(0,1)$ turning counterclockwise. You can probably prove that $x(\theta)=(\cos\theta,\sin\theta)$ (but I will NOT use this equality in the following).
Then if $f(\theta)= \|x(\theta+\tau)-x(\theta)\|$ with $\tau$ fixed, we have
$$ f'(\theta) = \frac{\langle x(\theta+\tau)-x(\theta),x'(\theta+\tau)-x'(\theta)\rangle}{\|x(\theta+\tau)-x(\theta)\|}=0$$
where $\langle\rangle$ is the usual scalar product. The above equality holds because 


*

*we have the equality $1=\|x(\theta\|$ which by differentiation gives us $\langle x'(\theta),x(\theta)\rangle=0$ 

*$\theta \mapsto \langle x(\theta+\tau),x(\theta)\rangle$ is constant as the scalar product of two vectors with constant angle $\tau$ between them. Differentiating we get
$$ \langle x(\theta+\tau),x'(\theta)\rangle+ \langle x'(\theta+\tau),x(\theta)\rangle =0$$
(this is again geometry)
Therefore $f$ is constant and $f(\theta)=f(0)$ which is what you want.
