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. For \( \cos 200^\circ \): Notice that: \[ 200^\circ = 180^\circ + 20^\circ. \] Using the cosine of a sum: \[ \cos(180^\circ + \theta) = -\cos \theta, \] we have: \[ \cos 200^\circ = -\cos 20^\circ = -p. \] 3. For \( \sin(-70^\circ) \): The sine function is odd, which means: \[ \sin(-\theta) = -\sin \theta. \] Therefore: \[ \sin(-70^\circ) = -\sin 70^\circ. \] Next, note that: \[ \sin 70^\circ = \cos 20^\circ = p, \] since: \[ \sin 70^\circ = \sin(90^\circ - 20^\circ) = \cos 20^\circ. \] Thus: \[ \sin(-70^\circ) = -p. \] 4. For \( \sin 10^\circ \): We use the half-angle formula for sine: \[ \sin^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{2}, \] with \(\theta = 20^\circ\), so that: \[ \sin^2 10^\circ = \frac{1 - \cos 20^\circ}{2}. \] Since \( \sin 10^\circ \) is positive (as \(10^\circ\) is in the first quadrant), we have: \[ \sin 10^\circ = \sqrt{\frac{1-\cos 20^\circ}{2}}. \] Substituting \( \cos 20^\circ = p \) gives: \[ \sin 10^\circ = \sqrt{\frac{1-p}{2}}. \] Thus, the answers are: - \( \cos 200^\circ = -p \). - \( \sin(-70^\circ) = -p \). - \( \sin 10^\circ = \sqrt{\frac{1-p}{2}} \).