# Proving the Sine Law using dot product.

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?

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

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 $$\alpha+\gamma+\beta=0.$$
• 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? May 25, 2021 at 17:49
• Try replacing $\beta=-(\alpha +\gamma).$ @Jacob May 25, 2021 at 18:20