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Suppose we have a triangle in $\mathbb{R}^2$ with distinct vertices $A,B,C$ joined by sides with length $a=\rm{dist}(B,C)$, $b=\rm{dist}(A,C)$, $c=\rm{dist}(A,B)$ and have angles $\alpha, \beta, \gamma$ at A,B,C respectively. Using only properties of the dot product is it possible to prove the sine law, that is $$\frac{\sin \alpha}{a}=\frac{\sin \beta}{b}=\frac{\sin \gamma}{c}$$ I have begun by rewriting $\frac{\sin \alpha}{a}$ in terms of distances and dot products, leaving me with

$$\frac{1}{\rm{dist}(B,C)}\sqrt{\left(1-\left(\frac{(C-A)\cdot(A-B)}{\rm{dist}(A,B)\rm{dist}(A,C)}\right)^2\right)}$$ Using the fact that $\cos \alpha= \frac{(C-A)\cdot(A-B)}{\rm{dist}(A,B)\rm{dist}(A,C)}$ and $\sin (\cos^{-1} (x)) = \sqrt{1-x^2}$. Is this a correct way to start because now it seems to me like I'm stuck?

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    $\begingroup$ Hint: $\operatorname{dist}(A,B)^2=(A-B)\cdot(A-B)$ $\endgroup$
    – Thomas Andrews
    May 25, 2021 at 17:11

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Use $d$ for $\operatorname{dist},$ and use that $$d(X,Y)^2=(X-Y)\cdot(X-Y).$$ Letting $\alpha=B-C,\beta =C-A,\gamma =A-B,$ Then your formula is:

$$\frac{\sqrt{(\gamma\cdot \gamma)(\beta \cdot \beta)-(\beta \cdot \gamma)^2}} {d(B,C)d(A,C)d(A,B)} $$

Expand what is inside the square root. Use $\alpha+\gamma+\beta=0.$

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  • $\begingroup$ Could you elaborate on how to expand the square root? I've just gone in a loop and done $(\gamma \cdot \gamma)=c^2, (\beta \cdot \beta) = b^2, (\beta \cdot \gamma) = bc \cos \alpha$ and that simply makes it reduce back to what I started with? $\endgroup$
    – Jacob
    May 25, 2021 at 17:49
  • $\begingroup$ Try replacing $\beta=-(\alpha +\gamma).$ @Jacob $\endgroup$
    – Thomas Andrews
    May 25, 2021 at 18:20

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