Application of Dot and Cross Products in Plane Trigonometry

Dot and cross products give slick, purely algebraic proofs of several plane-trigonometry results, with usual triangle $ABC$ notation ($a=BC,\,b=CA,\,c=AB$).

Cosine rule (Example 6.1). Since $\vec{BC}+\vec{CA}+\vec{AB}=\vec 0$, we get $\vec{BC}=-\vec{CA}-\vec{AB}$; dotting $\vec{BC}$ with itself and using $\vec{CA}\cdot\vec{AB}=|CA||AB|\cos(\pi-A)$ (the angle between the vectors $\vec{CA}$ and $\vec{AB}$, as drawn head-to-tail, is $\pi-A$) gives $a^2=b^2+c^2-2bc\cos A$, and cyclically for $B,C$.

Projection formula (Example 6.2): $a=b\cos C+c\cos B$, and cyclically — obtained the same way, dotting $\vec{BC}$ with itself but grouping differently.

Compound-angle identities (Examples 6.3 & 6.5, Exercise questions 9–10). Let $\hat a=\cos\alpha\,\hat i+\sin\alpha\,\hat j$ and $\hat b=\cos\beta\,\hat i+\sin\beta\,\hat j$ be unit vectors at angles $\alpha,\beta$ to the positive $x$-axis. Then:

• $\hat a\cdot\hat b=\cos\alpha\cos\beta+\sin\alpha\sin\beta$ is also $|\hat a||\hat b|\cos(\alpha-\beta)=\cos(\alpha-\beta)$, giving $\boxed{\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta}$. Replacing $\hat b$ by the unit vector at angle $-\beta$ (i.e. $\cos\beta\,\hat i-\sin\beta\,\hat j$) and redoing the dot product gives $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$.
• Taking $\hat a$ at angle $-\alpha$ (below the axis) and $\hat b$ at angle $\beta$ (above), $\hat a\times\hat b=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)\hat k$; since the angle swept from $\hat a$ to $\hat b$ is $\alpha+\beta$, $|\hat a\times\hat b|=\sin(\alpha+\beta)$, giving $\boxed{\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta}$. The companion $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$ follows the same way with both unit vectors above the axis. …