Given that \( \cos 20^{\circ}=p \) Without using a calculator, write EACH of the following in terms \( p \) : \( 5.2 .1 \quad \cos 200^{\circ} \) \( 5.2 .2 \quad \sin \left(-70^{\circ}\right) \) \( 5.2 .3 \quad \sin 10^{\circ} \)

1. We are given that \(\cos 20^\circ = p\). 2. To express each trigonometric function in terms of \(p\), we proceed as follows. **(a) \(\cos 200^\circ\)** Notice that \[ \cos 200^\circ = \cos (180^\circ + 20^\circ). \] Using the cosine addition formula: \[ \cos (180^\circ + \theta) = -\cos \theta, \] we have \[ \cos 200^\circ = -\cos 20^\circ = -p. \] **(b) \(\sin (-70^\circ)\)** Recall that sine is an odd function, so: \[ \sin (-\theta) = -\sin \theta. \] Thus, \[ \sin (-70^\circ) = -\sin 70^\circ. \] Recognize that \[ \sin 70^\circ = \cos (90^\circ - 70^\circ) = \cos 20^\circ = p. \] Hence, \[ \sin (-70^\circ) = -p. \] **(c) \(\sin 10^\circ\)** We use the cosine double-angle formula: \[ \cos 20^\circ = 1 - 2\sin^2 10^\circ. \] Substitute \(p\) for \(\cos 20^\circ\): \[ p = 1 - 2\sin^2 10^\circ. \] Solving for \(\sin^2 10^\circ\): \[ 2\sin^2 10^\circ = 1 - p \quad \Longrightarrow \quad \sin^2 10^\circ = \frac{1-p}{2}. \] Since \(10^\circ\) is acute, \(\sin 10^\circ\) is positive. Therefore, \[ \sin 10^\circ = \sqrt{\frac{1-p}{2}}. \] 3. The final answers in terms of \(p\) are: - \(\cos 200^\circ = -p\), - \(\sin (-70^\circ) = -p\), - \(\sin 10^\circ = \sqrt{\frac{1-p}{2}}\).