Answer
Using the trigonometric identity:
$\cos \left( {\theta - \psi } \right) = \cos \theta \cos \psi + \sin \theta \sin \psi $,
we obtain
${{\bf{e}}_\theta }\cdot{{\bf{e}}_\psi } = \cos \left( {\theta - \psi } \right)$
Work Step by Step
For angle $\theta$, we have ${{\bf{e}}_\theta } = \left( {\cos \theta ,\sin \theta } \right)$. And for angle $\psi$, we have ${{\bf{e}}_\psi } = \left( {\cos \psi ,\sin \psi } \right)$.
Using the definition of dot product, we get
${{\bf{e}}_\theta }\cdot{{\bf{e}}_\psi } = \left( {\cos \theta ,\sin \theta } \right)\cdot\left( {\cos \psi ,\sin \psi } \right)$
${{\bf{e}}_\theta }\cdot{{\bf{e}}_\psi } = \cos \theta \cos \psi + \sin \theta \sin \psi $
From trigonometry, we have
$\cos \left( {\theta - \psi } \right) = \cos \theta \cos \psi + \sin \theta \sin \psi $.
Therefore,
${{\bf{e}}_\theta }\cdot{{\bf{e}}_\psi } = \cos \left( {\theta - \psi } \right)$
