Elementary Linear Algebra 7th Edition

$\pi-\arccos\left( \frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\right)$.
The angle between the vectors $$\mathbf{u}=\left(\cos \frac{\pi}{6}, \sin \frac{\pi}{6}\right), \quad \mathbf{v}=\left(\cos \frac{3 \pi}{4}, \sin \frac{3 \pi}{4}\right)$$ is given by $$\arccos\left( \frac{u\cdot v}{\|u\| \|v\|}\right)=\pi-\arccos\left( \frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\right)$$