[proofplan] We prove the identity separately in the three constant-curvature models. In the Euclidean case, we place the vertex $\bar{p}$ at the origin and compute the squared distance between the two endpoint vectors. In positive curvature, we use the round sphere of radius $1/\sqrt{k}$ embedded in $\mathbb{R}^3$, where geodesic distance is encoded by the ambient Euclidean [inner product](/page/Inner%20Product). In negative curvature, we use the hyperboloid model in Minkowski space, where geodesic distance is encoded by the Lorentzian inner product; evaluating the two endpoint vectors gives the hyperbolic cosine law. [/proofplan] [step:Compute the Euclidean case by expanding the endpoint difference] Assume $k = 0$. Then $M_0^2$ is isometric to $\mathbb{R}^2$ with the Euclidean metric. Use an isometry to place $\bar{p}$ at $0 \in \mathbb{R}^2$, and choose the standard [orthonormal basis](/page/Orthonormal%20Basis) vectors \begin{align*} e_1 &= (1,0),& e_2 &= (0,1). \end{align*} Since $c = d(\bar{p},\bar{q})$ and $b = d(\bar{p},\bar{r})$, and the angle at $\bar{p}$ is $\alpha$, we may write \begin{align*} \bar{q} &= c e_1,\\ \bar{r} &= b(\cos \alpha\, e_1 + \sin \alpha\, e_2). \end{align*} Therefore \begin{align*} a^2 &= |\bar{r}-\bar{q}|^2\\ &= |b(\cos \alpha\, e_1 + \sin \alpha\, e_2) - c e_1|^2\\ &= |(b\cos\alpha-c)e_1 + b\sin\alpha\,e_2|^2\\ &= (b\cos\alpha-c)^2 + b^2\sin^2\alpha\\ &= b^2\cos^2\alpha - 2bc\cos\alpha + c^2 + b^2\sin^2\alpha\\ &= b^2 + c^2 - 2bc\cos\alpha. \end{align*} This proves the stated formula when $k=0$. [/step] [step:Represent the positive curvature model inside Euclidean space] Assume $k>0$, and define the radius \begin{align*} \rho := \frac{1}{\sqrt{k}}. \end{align*} Realize $M_k^2$ as the round sphere \begin{align*} S_\rho^2 := \{x \in \mathbb{R}^3 : x \cdot x = \rho^2\} \end{align*} with the metric induced by the Euclidean inner product $x \cdot y$ on $\mathbb{R}^3$. The geodesic distance on $S_\rho^2$ satisfies \begin{align*} x \cdot y = \rho^2 \cos\left(\frac{d(x,y)}{\rho}\right) \end{align*} for any two points $x,y \in S_\rho^2$ joined by a minimizing geodesic segment. Use an isometry of $S_\rho^2$ to place \begin{align*} \bar{p} = (\rho,0,0). \end{align*} Let \begin{align*} u &:= (0,1,0),& v &:= (0,\cos\alpha,\sin\alpha). \end{align*} Then $u,v \in T_{\bar{p}}S_\rho^2$, $|u|=|v|=1$, and $u \cdot v = \cos\alpha$. The unit-speed geodesic on $S_\rho^2$ starting at $\bar{p}$ with initial velocity $w \in T_{\bar{p}}S_\rho^2$ and $|w|=1$ is the map \begin{align*} \gamma_w: \mathbb{R} &\to S_\rho^2\\ t &\mapsto \cos\left(\frac{t}{\rho}\right)\bar{p} + \rho \sin\left(\frac{t}{\rho}\right)w. \end{align*} Thus, choosing the geodesic to $\bar{q}$ in direction $u$ and the geodesic to $\bar{r}$ in direction $v$, we have \begin{align*} \bar{q} &= \cos\left(\frac{c}{\rho}\right)\bar{p} + \rho \sin\left(\frac{c}{\rho}\right)u,\\ \bar{r} &= \cos\left(\frac{b}{\rho}\right)\bar{p} + \rho \sin\left(\frac{b}{\rho}\right)v. \end{align*} [guided] The useful feature of the sphere model is that geodesic distance can be recovered from the ordinary Euclidean inner product in $\mathbb{R}^3$. We set \begin{align*} \rho := \frac{1}{\sqrt{k}}, \end{align*} so the model space of curvature $k$ is the round sphere \begin{align*} S_\rho^2 := \{x \in \mathbb{R}^3 : x \cdot x = \rho^2\}. \end{align*} For two points $x,y \in S_\rho^2$ connected by a minimizing geodesic segment, the central angle between the radius vectors $x$ and $y$ is $d(x,y)/\rho$, hence \begin{align*} x \cdot y = \rho^2 \cos\left(\frac{d(x,y)}{\rho}\right). \end{align*} We may rotate the sphere without changing distances or angles, so place \begin{align*} \bar{p} = (\rho,0,0). \end{align*} The tangent plane at $\bar{p}$ is the plane perpendicular to $\bar{p}$, namely \begin{align*} T_{\bar{p}}S_\rho^2 = \{0\}\times \mathbb{R}^2. \end{align*} Choose unit tangent vectors \begin{align*} u &:= (0,1,0),& v &:= (0,\cos\alpha,\sin\alpha). \end{align*} They satisfy \begin{align*} u \cdot v = \cos\alpha, \end{align*} so the angle between them is exactly $\alpha$. The unit-speed geodesic starting at $\bar{p}$ in a unit tangent direction $w \in T_{\bar{p}}S_\rho^2$ is \begin{align*} \gamma_w: \mathbb{R} &\to S_\rho^2\\ t &\mapsto \cos\left(\frac{t}{\rho}\right)\bar{p} + \rho \sin\left(\frac{t}{\rho}\right)w. \end{align*} This formula stays on $S_\rho^2$ because $\bar{p}\cdot w=0$, $|\bar{p}|=\rho$, and $|w|=1$. It has speed one because differentiating gives \begin{align*} \gamma_w'(t) = -\frac{1}{\rho}\sin\left(\frac{t}{\rho}\right)\bar{p} + \cos\left(\frac{t}{\rho}\right)w, \end{align*} and therefore \begin{align*} |\gamma_w'(t)|^2 = \sin^2\left(\frac{t}{\rho}\right) + \cos^2\left(\frac{t}{\rho}\right) = 1. \end{align*} Hence the points at distances $c$ and $b$ from $\bar{p}$ in the two chosen directions are \begin{align*} \bar{q} &= \cos\left(\frac{c}{\rho}\right)\bar{p} + \rho \sin\left(\frac{c}{\rho}\right)u,\\ \bar{r} &= \cos\left(\frac{b}{\rho}\right)\bar{p} + \rho \sin\left(\frac{b}{\rho}\right)v. \end{align*} [/guided] [/step] [step:Evaluate the Euclidean inner product to obtain the spherical formula] Using $\bar{p}\cdot u=\bar{p}\cdot v=0$, $\bar{p}\cdot\bar{p}=\rho^2$, and $u\cdot v=\cos\alpha$, we compute \begin{align*} \bar{q}\cdot\bar{r} &= \rho^2\cos\left(\frac{c}{\rho}\right)\cos\left(\frac{b}{\rho}\right) + \rho^2\sin\left(\frac{c}{\rho}\right)\sin\left(\frac{b}{\rho}\right)\cos\alpha. \end{align*} On the other hand, since $a=d(\bar{q},\bar{r})$, \begin{align*} \bar{q}\cdot\bar{r} = \rho^2\cos\left(\frac{a}{\rho}\right). \end{align*} Dividing by $\rho^2>0$ gives \begin{align*} \cos\left(\frac{a}{\rho}\right) = \cos\left(\frac{b}{\rho}\right)\cos\left(\frac{c}{\rho}\right) + \sin\left(\frac{b}{\rho}\right)\sin\left(\frac{c}{\rho}\right)\cos\alpha. \end{align*} Since $1/\rho=\sqrt{k}$, this is \begin{align*} \cos(\sqrt{k}\,a) = \cos(\sqrt{k}\,b)\cos(\sqrt{k}\,c) + \sin(\sqrt{k}\,b)\sin(\sqrt{k}\,c)\cos\alpha. \end{align*} This proves the stated formula when $k>0$. [/step] [step:Represent the negative curvature model inside Minkowski space] Assume $k<0$, and define \begin{align*} \rho := \frac{1}{\sqrt{-k}}. \end{align*} Let $\mathbb{R}^{2,1}$ denote $\mathbb{R}^3$ equipped with the Lorentzian [bilinear form](/page/Bilinear%20Form) \begin{align*} \langle x,y\rangle_L := -x_0y_0 + x_1y_1 + x_2y_2 \end{align*} for $x=(x_0,x_1,x_2)$ and $y=(y_0,y_1,y_2)$. Realize $M_k^2$ as the upper hyperboloid \begin{align*} H_\rho^2 := \{x \in \mathbb{R}^{2,1} : \langle x,x\rangle_L = -\rho^2,\ x_0>0\} \end{align*} with its induced Riemannian metric. The distance on $H_\rho^2$ satisfies \begin{align*} -\langle x,y\rangle_L = \rho^2\cosh\left(\frac{d(x,y)}{\rho}\right) \end{align*} for all $x,y \in H_\rho^2$. Use an isometry of $H_\rho^2$ to place \begin{align*} \bar{p} = (\rho,0,0). \end{align*} Let \begin{align*} u &:= (0,1,0),& v &:= (0,\cos\alpha,\sin\alpha). \end{align*} Then $u,v \in T_{\bar{p}}H_\rho^2$, $\langle u,u\rangle_L=\langle v,v\rangle_L=1$, and $\langle u,v\rangle_L=\cos\alpha$. The unit-speed geodesic on $H_\rho^2$ starting at $\bar{p}$ with initial velocity $w \in T_{\bar{p}}H_\rho^2$ and $\langle w,w\rangle_L=1$ is the map \begin{align*} \eta_w: \mathbb{R} &\to H_\rho^2\\ t &\mapsto \cosh\left(\frac{t}{\rho}\right)\bar{p} + \rho\sinh\left(\frac{t}{\rho}\right)w. \end{align*} Thus \begin{align*} \bar{q} &= \cosh\left(\frac{c}{\rho}\right)\bar{p} + \rho\sinh\left(\frac{c}{\rho}\right)u,\\ \bar{r} &= \cosh\left(\frac{b}{\rho}\right)\bar{p} + \rho\sinh\left(\frac{b}{\rho}\right)v. \end{align*} [/step] [step:Evaluate the Lorentzian inner product to obtain the hyperbolic formula] Using $\langle \bar{p},u\rangle_L=\langle \bar{p},v\rangle_L=0$, $\langle \bar{p},\bar{p}\rangle_L=-\rho^2$, and $\langle u,v\rangle_L=\cos\alpha$, we compute \begin{align*} \langle \bar{q},\bar{r}\rangle_L &= -\rho^2\cosh\left(\frac{c}{\rho}\right)\cosh\left(\frac{b}{\rho}\right) + \rho^2\sinh\left(\frac{c}{\rho}\right)\sinh\left(\frac{b}{\rho}\right)\cos\alpha. \end{align*} Since $a=d(\bar{q},\bar{r})$, the hyperboloid distance relation gives \begin{align*} -\langle \bar{q},\bar{r}\rangle_L = \rho^2\cosh\left(\frac{a}{\rho}\right). \end{align*} Combining the last two identities and dividing by $\rho^2>0$ yields \begin{align*} \cosh\left(\frac{a}{\rho}\right) = \cosh\left(\frac{b}{\rho}\right)\cosh\left(\frac{c}{\rho}\right) - \sinh\left(\frac{b}{\rho}\right)\sinh\left(\frac{c}{\rho}\right)\cos\alpha. \end{align*} Since $1/\rho=\sqrt{-k}$, this is \begin{align*} \cosh(\sqrt{-k}\,a) = \cosh(\sqrt{-k}\,b)\cosh(\sqrt{-k}\,c) - \sinh(\sqrt{-k}\,b)\sinh(\sqrt{-k}\,c)\cos\alpha. \end{align*} This proves the stated formula when $k<0$. Together with the Euclidean and spherical cases, the theorem follows for every $k \in \mathbb{R}$. [/step]