Consider two unit vectors a and b, as shown in Fig. 1-19.
Find the trigonometric relations for the sine and cosine of the sum and diference of two angles from the values of $\mathbf{a} \cdot \mathbf{b}$ and $\mathbf{a} \times \mathbf{b}$.
Solution: First,
$$
\begin{aligned}
&\mathbf{a}=\cos \alpha \mathbf{i}+\sin \alpha \mathbf{j} \\
&\mathbf{b}=\cos \beta \mathbf{i}+\sin \beta \mathbf{j}
\end{aligned}
$$
Now we can write the product $\mathbf{a} \cdot \mathbf{b}$ in two different ways. From Eq. 1-11,
$$
\mathbf{a} \cdot \mathbf{b}=\cos \alpha \cos \beta+\sin \alpha \sin \beta
$$
and, from Eq. 1-5,
$$
\mathbf{a} \cdot \mathbf{b}=\cos (\alpha-\beta)
$$
Thus
$$
\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta
$$
If we set $\beta^{\prime}=-\beta$
$$
\cos \left(\alpha+\beta^{\prime}\right)=\cos \alpha \cos \beta^{\prime}-\sin \alpha \sin \beta^{\prime}
$$
Similarly, from Eq. 1-22,
$$
\begin{aligned}
\mathbf{a} \times \mathbf{b}=& \mid \begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\cos \alpha & \sin \alpha & 0 \\
\cos \beta & \sin \beta & 0
\end{array} \\
&=(\cos x \sin \beta-\cos \beta \sin \alpha) \mathbf{k}
\end{aligned}
$$
and, from Eqs. $1-14$ and $1-15$
$$
\mathbf{a} \times \mathbf{b}=-\sin (\alpha-\beta) \mathbf{k}
$$
Thus
$$
\sin (\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta
$$
Setting again $\beta^{t}=-\beta$
$$
\sin \left(\alpha+\beta^{\prime}\right)=\sin \alpha \cos \beta^{\prime}+\cos \alpha \sin \beta^{t}
$$