I'm trying to prove that given 2 vectors $\vec{a} = A(\cos{\alpha}, \sin{\alpha}) $ and $\vec{b} = B(\cos{\beta}, \sin{\beta}) $ the following relation is true by using the exterior product with the components of each vector:

$$|\sin{(\alpha - \beta)}| = |\sin{\alpha} \cos{\beta} - \cos{\alpha}\sin{\beta}|$$

My approach is that since we know $|\vec{a} \times \vec{b} |= |\vec{a}| |\vec{b}|\sin{\theta}$ we can follow from there but I'm stuck. I can't figure how to introduce $\alpha$ nor $\beta$. Also, which component of these vectors should I use or is there an approach without necessarily using the components. Thanks.


Since you want to calculate the cross product of $\vec{a}$ and $\vec{b}$, you are supposed to extend it to 3-D cases. That is, rewrite $$\vec{a}=(A\cos\alpha,A\sin\alpha,0)\quad{\rm and}\quad\vec{b}=(B\cos\beta,B\sin\beta,0).$$

Then the exterior product of $\vec{a}$ and $\vec{b}$ is $$\vec{a}\times\vec{b}=(0,0,A\cos\alpha\cdot B\sin\beta-A\sin\alpha\cdot B\cos\beta).$$ You can refer to this link for the formula.

As a result, $$|\vec{a}\times\vec{b}|=|AB|\cdot|\cos\alpha\sin\beta-\sin\alpha\cos\beta|=|AB|\cdot|\sin\alpha\cos\beta-\cos\alpha\sin\beta|.$$ On the other hand, we have $|\vec{a}|=|A|$, $|\vec{b}|=|B|$, the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is exactly $\beta-\alpha$. Applying the formula $|\vec{a} \times \vec{b} |= |\vec{a}| |\vec{b}|\sin{\theta}$, we have $$|AB|\cdot|\sin\alpha\cos\beta-\cos\alpha\sin\beta|=|A||B|\sin(\beta-\alpha). $$ By cancelling the common coefficients and taking the absolute value of each sides, we can obtain the desired formula.

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