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

To express \( \cos 200^{\circ} \) in terms of \( p \), we start by recognizing that \( 200^{\circ} = 180^{\circ} + 20^{\circ} \). We can use the cosine identity for angles greater than 180 degrees: \[ \cos(180^{\circ} + \theta) = -\cos(\theta). \] So, we have: \[ \cos 200^{\circ} = \cos(180^{\circ} + 20^{\circ}) = -\cos 20^{\circ} = -p. \] Next, to find \( \sin(-70^{\circ}) \), we can use the odd function property of sine: \[ \sin(-\theta) = -\sin(\theta). \] Thus: \[ \sin(-70^{\circ}) = -\sin(70^{\circ}). \] Now using the co-function identity: \[ \sin(90^{\circ} - \theta) = \cos(\theta), \] we find: \[ \sin(70^{\circ}) = \sin(90^{\circ} - 20^{\circ}) = \cos(20^{\circ}). \] Therefore: \[ \sin(-70^{\circ}) = -\sin(70^{\circ}) = -\cos(20^{\circ}) = -p. \] For \( \sin 10^{\circ} \), we can use the angle subtraction formula: \[ \sin(10^{\circ}) = \sin(30^{\circ} - 20^{\circ}). \] Using the sine subtraction formula \(\sin(a - b) = \sin a \cos b - \cos a \sin b\), we have: \[ \sin(30^{\circ}) = \frac{1}{2}, \quad \cos(30^{\circ}) = \frac{\sqrt{3}}{2}, \quad \sin(20^{\circ}) = \sqrt{1 - \cos^2(20^{\circ})} = \sqrt{1 - p^2}. \] Thus: \[ \sin(10^{\circ}) = \sin(30^{\circ}) \cos(20^{\circ}) - \cos(30^{\circ}) \sin(20^{\circ}) = \frac{1}{2} p - \frac{\sqrt{3}}{2} \sqrt{1 - p^2}. \] Putting this all together: - \( \cos 200^{\circ} = -p \) - \( \sin(-70^{\circ}) = -p \) - \( \sin 10^{\circ} = \frac{1}{2} p - \frac{\sqrt{3}}{2} \sqrt{1 - p^2} \)