# Statement(s) > [!NOTE] Statement 1 (Spherical metric satisfies triangle inequality) > Let $P,Q,R\in S_{n}$, the [[n-Sphere|unit n-Sphere]] with $\alpha:=d(Q,R),\quad \beta:= d(P,Q),\quad \gamma:=d(P,R),$where $d$ denotes the [[Spherical Metric|spherical metric]], then $\alpha \leq \beta+ \gamma$with equality, or with $\alpha+\beta+\gamma=2\pi$ iff $a=\pi,$ where $a$ is the [[Spherical Angle|spherical angle]] between [[Spherical Line|lines]] $PQ$ and $PR.$ Furthermore, $a=0$ iff $\alpha=|\beta-\gamma|.$ # Proof(s) ###### Proof of statement 1 (MA243): Points $P,Q,R$ are contained in a vector space $V$ of dimension 3, and so lie on the unit sphere $S^2\cong V\cap S^n.$ Hence, it is enough to prove the result for $n=2.$ We may assume $P,Q,R$ are distinct, since otherwise the result is easy. By definition, $\alpha,\beta,\gamma\in[0,\pi]$, so $\sin\alpha,\sin\beta\geq0.$ Since $\cos a\geq-1$, it follows from [[Main Spherical Trigonometric Formula]] that $\tag{1}\begin{aligned}\cos\alpha&=\cos\beta\cdot\cos\gamma+\sin\beta\cdot\sin\gamma\cdot\cos a\\&\geq\cos\beta\cdot\cos\gamma-\sin\beta\cdot\sin\gamma=\cos(\beta+\gamma)\end{aligned}$ Case $\beta+\gamma\leq\pi{:}$ By definition, $\alpha\in[0,\pi].$ On $[0,\pi],\cos(x)$ is strictly decreasing, so $\cos\alpha\geq\cos(\beta+\gamma)\Rightarrow\alpha\leq\beta+\gamma.$ There is equality in $(1)$ iff $\cos(a)=-1,$ which is if and only if $a=\pi.$ Case $\beta+\gamma>\pi{:}$ Since $\alpha\in[0,\pi],$ the triangle inequality is immediate. Equality in $(1)$ occurs iff $a=\pi,$ but now $\beta+\gamma>\pi$ so $\cos(\beta+\gamma)=\cos(\alpha)\iff\alpha=2\pi=(\beta+\gamma)\iff\alpha+\beta+\gamma=2\pi.$ The final statement follows similarly since $a=0\iff \cos(\alpha)=\cos(\beta-\gamma).$ $\blacksquare$ # Application(s) **Consequences**: **Examples**: # Bibliography