## Precalculus (10th Edition)

$1$
The identity says: $\sin^2(\theta)+\cos^2(\theta)=1$. We also know that $\cos(\theta)=\sin(90^{\circ}-\theta)$. Hence, $\cos^2(40^{\circ})+\cos^2(50^{\circ}) \\=\cos^2(40^{\circ})+\sin^2(90^{\circ}-50^{\circ})\\ =\cos^2(40^{\circ})+\sin^2(40^{\circ}) \\=1$