Using vectors, prove sin(α-β)=sinαcosβ-sinβcosα
Using vectors, prove $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$.
Explanation of Method
Vector proof of the sine subtraction formula.
Step 1: Two unit vectors
\[ \vec a=\{\cos\alpha,\ \sin\alpha,\ 0\},\qquad \vec b=\{\cos\beta,\ \sin\beta,\ 0\}. \]
Step 2: $\vec b\times\vec a$ in two ways
Geometrically: the rotation from $\vec b$ to $\vec a$ is counter-clockwise by $\alpha-\beta$ (seen from $+z$). By the right-hand rule and $|\vec a|=|\vec b|=1$:
\[ \vec b\times\vec a=\sin(\alpha-\beta)\,\vec k. \]
(For $\alpha-\beta$ negative or beyond $180^\circ$, the signed sine automatically gives the correct direction.)
In coordinates:
\[ \vec b\times\vec a=\begin{vmatrix}\vec i&\vec j&\vec k\\\cos\beta&\sin\beta&0\\\cos\alpha&\sin\alpha&0\end{vmatrix}=(\cos\beta\sin\alpha-\sin\beta\cos\alpha)\,\vec k. \]
Step 3: Comparison
\[\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha\qquad\blacksquare\]
Companion formula: $\vec a\cdot\vec b=\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$ (Problem 64).
Result
\[\boxed{\,\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha\qquad\blacksquare\,}\]
Remark
Correction. The source writes $\vec a\times\vec b=\sin(\alpha-\beta)\vec k$. The coordinates then give $\sin(\alpha-\beta)=\cos\alpha\sin\beta-\sin\alpha\cos\beta$, which has the wrong sign, and the source "fixes" this by multiplying one side by $-1$. That step is not valid. The source's error is the direction: $\vec a\times\vec b=\sin(\beta-\alpha)\vec k$, because the turn goes from $\vec a$ to $\vec b$. Using $\vec b\times\vec a$, as above, gives the formula directly.