Given Sinlb \( =p \) Determine the following in term of \( p \) withait using a colculato Q 1. Sin \( 196^{\circ} \) 2. \( \operatorname{Cos} 16^{\circ} \)

To find \(\sin 196^\circ\), we can use the identity \(\sin(180^\circ + \theta) = -\sin \theta\). Here, \(196^\circ = 180^\circ + 16^\circ\), so \(\sin 196^\circ = -\sin 16^\circ\). Since \(\sin 16^\circ\) is given as \(p\), we have \(\sin 196^\circ = -p\). For \(\cos 16^\circ\), we can use the Pythagorean identity relating sine and cosine: \(\cos^2 \theta + \sin^2 \theta = 1\). Since \(\sin 16^\circ = p\), it follows that \(\cos^2 16^\circ = 1 - p^2\), which gives us \(\cos 16^\circ = \sqrt{1 - p^2}\). So, the answers are: 1. \(\sin 196^\circ = -p\) 2. \(\cos 16^\circ = \sqrt{1 - p^2}\)