Prove angle addition holds for $\mathbb{R}^n$

Question

Define $\theta(u,v)=\cos^{-1}(\frac{u\cdot v}{|u||v|})$ be the angle between $u,v\in \mathbb{R}^n$, where $u\cdot v$ is the standard inner product and $|x|=\sqrt{x\cdot x}$ for all $x\in \mathbb{R}^n$. Let $w=su+tv$ where $s,t\geq 0$ are scalars. Show $\theta(u,v)=\theta(u,w)+\theta(v,w)$ for $u,v\neq 0$.

Attempts: We can show $\theta(u,v)=\theta(u,tv)$ thus we only need to show the statement when $|u|=|v|=1$. Moreover, we can find $a,b$ so that $(u-aw)\cdot w=0$ and $(v-bw)\cdot w=0$. That should simplify the question as well since we turn them into orthogonal.

Answer 1

Assume $|u|=|v|=1$ and $w=su+tv$, and denote $u\cdot v=x$. Then $\displaystyle |w|= \sqrt{s^2+t^2+2stx}$, $\displaystyle \cos\theta(u,w)= \frac{s+tx}{\sqrt{s^2+t^2+2stx}}$ and $\displaystyle \cos\theta(v,w)= \frac{t+sx}{\sqrt{s^2+t^2+2stx}}$. Also $\displaystyle \sin\theta(u,w)= \sqrt{1-\cos^2\theta(u,w)}= \sqrt{\frac{s^2+t^2+2stx-s^2-t^2x^2-2stx}{s^2+t^2+2stx}}= \frac{t\sqrt{1-x^2}}{\sqrt{s^2+t^2+2stx}}$ and similarly $\displaystyle \sin\theta(v,w)= \frac{s\sqrt{1-x^2}}{\sqrt{s^2+t^2+2stx}}$. Now $\displaystyle \cos(\theta(u,w)+\theta(v,w))= \cos\theta(u,w)\cos\theta(v,w)- \sin\theta(u,w)\sin\theta(v,w)=$ $$=\frac{st+s^2x+t^2x+stx^2-st(1-x^2)}{s^2+t^2+2stx}= \frac{(s^2+t^2+2stx)x}{s^2+t^2+2stx}= x=\cos\theta(u,v).$$

Answer 2

Here's an alternative approach. We can replace $u$ by $u'$ and $v$ by $v'$ where \begin{align} u' &= \frac{u}{ \| u \|} \\ v' &= \frac{v - (v \cdot u)u}{\|v - (v \cdot u)u\|} \end{align} Letting $c = \|v - (v \cdot u)u\|$, we can see that $$ a' u' + b' v' = au + bv $$ where $$ \pmatrix{a\\b} = \pmatrix{\frac{1}{\|u\|}& -\frac{v \cdot u}{c}\\0&\frac{1}{c}} \pmatrix{a' \\b'} $$ and because that $2 \times 2$ matrix is invertible, we can also express $a'$ and $b'$ in terms of $a$ and $b$. Hence any point of the form $su + tv$ is also of the form $s'u' + t'v'$, and vice versa, and we may a well assume that $u$ and $v$ are perpendicular unit vectors.

Now apply Gram-Schmidt to extend $u, v$ to an orthonormal basis of $\Bbb R^n$, $$ u, v, b_3, b_3, \ldots, b_n $$ and consider the matrix $M$ whose columns are those orthonormal vectors in order. The map $$ x \mapsto M^t x $$ is an isometry, and sends $u$ to $e_1$, $v$ to $e_2$, and $w = su + tv$ to the point $(s, t, 0, \ldots 0)$.

The problem is now reduced to showing that for a point $p = (s, t) \ne (0,0)$ in the plane, the angle from the ray in direction $e_1$ to the ray in direction $p$, plus the angle from that second ray to the ray in direction $e_2$, sum to $\pi/2$. But this is just an application of the law of cosines in the euclidean plane.

(The condition $w \ne (0,0)$ was missing from the original hypotheses, but should have been there.)

This is all pretty long and elaborate, but the short form is "consider the plane spanned by $u$ and $v$; there's an isometry to the standard euclidean plane; the law of cosines in the euclidean plane then proves the result."

Answer 3

Let $c: \mathbb{R} \longrightarrow \mathbb{R}^2, \ t \longmapsto (\cos(t), \sin(t))$ be a parametrization of the unit circle with center $0$ and let $t_1, t_2 \in \mathbb{R}$ such that $0 \leq t_2 - t_1 \leq \pi$. Define $U : = c(t_1), \ V:= c(t_2)$. Then we have

$$ \sphericalangle(U, 0, V) = \int_{t_1}^{t_2} \|\dot{c}(t) \| \ \mathrm{d}t. \tag{$\ast$} $$

Proof. We clearly have

$$\int_{t_1}^{t_2} \| \dot{c}(t) \| \ \mathrm{d}t = \int_{t_1}^{t_2} \sqrt{\sin^2(t) + \cos^2(t)}\ \mathrm{d}t = t_2 - t_1.$$ It is $U = (\cos(t_1), \sin(t_1)), \ V = (\cos(t_2), \sin(t_2))$ and therefore

$$\cos(\sphericalangle (U, 0, V)) = \frac{\langle U, V \rangle}{\|U \| \| V\|} = \cos(t_1)\cos(t_2) + \sin(t_1) \sin(t_2) = \cos(t_2 - t_1). $$

Since $t_2 - t_1 \in [0, \pi]$, $(\ast)$ follows.

Now let $t_1 \leq t_2 \leq t_3$ such that $0 \leq t_3 - t_1 \leq \pi$ and additionally $W = c(t_3)$. Then $(\ast)$ yields

$$\sphericalangle(U, 0, V) + \sphericalangle(V, 0, W) = \int_{t_1}^{t_2} \|\dot{c}(t) \| \ \mathrm{d}t + \int_{t_2}^{t_3} \|\dot{c}(t) \| \ \mathrm{d}t = \int_{t_1}^{t_3} \|\dot{c}(t) \| \ \mathrm{d}t = \sphericalangle(U, 0, W).$$