## Precalculus (6th Edition) Blitzer

$\cos{71^o}$
RECALL: The cofunction Identities: (1) $\sin{\theta} = \cos{(90^o-\theta)}$ (2) $\cos{\theta} = \sin{(90^o-\theta)}$ (3) $\tan{\theta} = \cot{(90^o-\theta)}$ (4) $\cot{\theta} = \tan{(90-\theta^o)}$ (5) $\csc{\theta} = \sec{(90^o-\theta)}$ (6) $\sec{\theta} = \csc{(90^0-\theta)}$ Use identity (1) to obtain: $\sin{19^o} = \cos{(90^o-19^o)} = \cos{71^o}$