I'm reading my textbook and for some reason, it does not present the proof for the spherical polar law of cosine which is:
$$ \cos(a)=\frac{\cos(A)+\cos(B)\cos(C)}{\sin(B) \sin(C)}$$
It does present the proof for spherical law of cosine which is:
$$ \cos(A)=\frac{\cos(a)-\cos(b)\cos(c)}{\sin(b) \sin(c)}$$
where $A$ is the angle and $a,b,c$ are the sides.
in which they use cross product and dot product where for example $|b\times c|=\sin a$ and $(b,c)=\cos a$. So i'm not sure but I believe that for the proof for the spherical polar law of cosine, it can use a similar method. However, like i said, i'm not sure. For example, does $|B\times C|=\sin A$ ?
If anyone can help start this proof or show me, either way that would be great.
Best Answer
Here is a different proof from Boris Springborn's lecture notes from the TU Berlin, which uses the Gram matrix to obtain the side cosine, angle cosine and four more equations by simultaneous permutations of $a,b,c$ and $\alpha^\prime, \beta^\prime,\gamma^\prime$.
Proof:
Let $V = \left(A \ B \ C\right) \in \mathbb{R}^{3 \times 3}$ be the matrix whose columns are the vertices of the spherical triangle, cosidered as column vectors. Then the Gram matrix for $A,B,C$ is $$ G = V^TV= \begin{pmatrix} \langle A,A\rangle & \langle A,B\rangle & \langle A,C\rangle \\ \langle B,A\rangle & \langle B,B\rangle & \langle B,C\rangle \\ \langle C,A\rangle & \langle C,B\rangle & \langle C,C\rangle \\ \end{pmatrix} = \begin{pmatrix} 1 & \cos c & \cos b \\ \cos c & 1 & \cos a \\ \cos b & \cos a & 1 \\ \end{pmatrix}. $$ (Note for later that $\mathrm{det}(G) > 0$). Similarly, let $W = \left( A^\prime \ B^\prime \ C^\prime \right)$ be the matrix of poles. Their Gram matrix is $$ G^\prime = W^TW = \begin{pmatrix} \langle A^\prime,A^\prime\rangle & \langle A^\prime,B^\prime\rangle & \langle A^\prime,C^\prime\rangle \\ \langle B^\prime,A^\prime\rangle & \langle B^\prime,B^\prime\rangle & \langle B^\prime,C^\prime\rangle \\ \langle C^\prime,A^\prime\rangle & \langle C^\prime,B^\prime\rangle & \langle C^\prime,C^\prime\rangle \\ \end{pmatrix} = \begin{pmatrix} 1 & \cos \gamma^\prime & \cos \beta^\prime \\ \cos \gamma^\prime& 1 & \cos \alpha^\prime \\ \cos \beta^\prime & \cos \alpha^\prime & 1 \\ \end{pmatrix}. $$. Also, $$ W^TV= \begin{pmatrix} \langle A^\prime,A\rangle & \langle A^\prime,B\rangle & \langle A^\prime,C\rangle \\ \langle B^\prime,A\rangle & \langle B^\prime,B\rangle & \langle B^\prime,C\rangle \\ \langle C^\prime,A\rangle & \langle C^\prime,B\rangle & \langle C^\prime,C\rangle \\ \end{pmatrix} = \begin{pmatrix} \langle A^\prime,A\rangle & 0 & 0 \\ 0 & \langle B^\prime,B\rangle & 0 \\ 0 & 0 & \langle C^\prime,C\rangle \\ \end{pmatrix} =\colon D. $$ is a diagonal matrix with positive entries. So $W^T = DV^{-1}$ and $W = (V^t)^{-1}D$, and $$ G^\prime = DV^{-1}(V^T)^{-1}D = D(V^TV)^{-1}D = DG^{-1}D. \quad (\star) $$ The inverse of $G$ is $$ G^{-1} = \frac{1}{\det(G)}\begin{pmatrix} \sin^2 a & -\cos c + \cos a \cos b & - \cos b + \cos c \cos a \\ -\cos c + \cos a \cos b & \sin^2 b & -\cos a + \cos b \cos c \\ - \cos b + \cos c \cos a & -\cos a + \cos b \cos c & \sin^2 c \end{pmatrix}. $$ Substitute this into $(\star)$ and consider diagonal elements:
One finds $1 = D_{11}^{2} \frac{1}{\mathrm{det}(G)}\sin^2 a$, therefore $D_{11} = \frac{\sqrt{\mathrm{det}(G)}}{\sin a}$, andsimilarly $D_{22} =\frac{\sqrt{\mathrm{det}(G)}}{\sin b}$,$D_{33} = \frac{\sqrt{\mathrm{det}(G)}}{\sin c}$. Now consider for example element $(3,2)$ in $(\star)$:$$ \cos \alpha^\prime = D_{33}\frac{1}{\mathrm{det}(G)}\left(-\cos a + \cos b \cos c \right)D_{22}. \qquad \text{(Convince yourself!)} $$ This is the side cosine theorem: $\cos \alpha^\prime = \frac{-\cos a + \cos b \cos c}{\sin b \sin c}$. The angle cosine theorem is the side cosine theorem applied to the polar triangle. Hence, $\cos a = \frac{-\cos \alpha^\prime + \cos \beta^\prime \cos \gamma^\prime}{\sin \beta^\prime \sin \gamma^\prime}$.