Answer
$\cos{71^o}$
Work Step by Step
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:
$\sin{19^o} = \cos{(90^o-19^o)} = \cos{71^o}$
