## Precalculus (6th Edition) Blitzer

$\sin^2 (\dfrac{\pi}{7}) +\cos^2 (\dfrac{\pi}{7}) =1$
Since, $\sin^2 \theta +\cos^2 \theta=1$ Here, $\theta = \dfrac{\pi}{7}$ Thus, we have $\sin^2 (\dfrac{\pi}{7}) +\cos^2 (\dfrac{\pi}{7}) =1$