Why is the ratio of the $||a-b||_2$ and $||a+b||_2$, $\tan[\angle (a,b)/2]$ when $a$ and $b$ are unit vectors with $a^Tb \geq 0$?

Question

Let $a$, $b\in\mathbb{R}^n$.
When $a^Tb \geq 0$ and $\|a\|_2=\|b\|_2=1$, can we prove $$\frac{\|a-b\|_2}{\|a+b\|_2} = \tan{\frac{\angle(a,b)}{2}}\text{?}$$

In the 2D case, this is clear from examining the parallelogram spanned by $a$ and $b$; compare this diagram by commenter across. But I can't visualize vectors in higher dimensions (say, $n=4$), so I'd like an algebraic proof. Is there one?

Answer 1

When $a^Tb\ge 0$ angles are acute, so we can assume $\angle (a+b,a) = \frac{\angle(a,b)}{2}$ in $\mathbb{R}^n$.

decompose $a+b$ along $a$ and perpendicular of $a$
say $x$ is the magnitude of component vector parallel to $a$: $$x=a\cdot(a+b) = 1+a\cdot b $$

say $y$ is the magnitude of component vector perpendicular to $a$: $$y^2 = (a+b)^2-x^2 = (1+a\cdot b)(1-a\cdot b)$$

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$$\tan\frac{\angle (a,b)}{2} = \frac{y}{x} =\sqrt{\frac{1-a\cdot b}{1+a\cdot b}} =\frac{\|a-b\|}{\|a+b\|}$$

Answer 2

Alternatively, without referring to angles directly in $\mathbb{R}^n$, use the trig identity:

$$\tan \frac{u}{2} = \sqrt{\frac{1-\cos u}{1+\cos u}}$$

note $\cos\angle (a,b) = a\cdot b $

$$\tan\frac{\angle (a,b)}{2} = \sqrt{\frac{1-a\cdot b}{1+a\cdot b}} = \frac{\|a-b\|}{\|a+b\|}$$