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

To express each of the trigonometric functions in terms of \( p = \cos 20^{\circ} \), let's convert the angles accordingly. 1. **For \( \cos 200^{\circ} \)**: \[ \cos 200^{\circ} = \cos (180^{\circ} + 20^{\circ}) = -\cos 20^{\circ} = -p \] 2. **For \( \sin (-70^{\circ}) \)**: \[ \sin (-70^{\circ}) = -\sin 70^{\circ} = -\sin (90^{\circ} - 20^{\circ}) = -\cos 20^{\circ} = -p \] 3. **For \( \sin 10^{\circ} \)**: Using the sine complement identity: \[ \sin 10^{\circ} = \sin (90^{\circ} - 80^{\circ}) = \cos 80^{\circ} \] We can further relate this back to \( p \) since \( \cos 80^{\circ} = \sin 10^{\circ} \) and is equal to \( \sin (90^{\circ} - 20^{\circ}) = \sin 70^{\circ} \). To express it in terms of \( p \), we can find that: \[ \sin 70^{\circ} = \cos 20^{\circ} = p \] So, \[ \sin 10^{\circ} = \sin (90^{\circ} - 80^{\circ}) = \cos 80^{\circ} = \cos(90^{\circ} - 70^{\circ}) = \sin 70^{\circ} = p \] But again, we have: \[ \sin 10^{\circ} = \sin (20^{\circ} \cdot 0.5) = \frac{1}{2} \cdot \sqrt{1 - \cos^2 20^{\circ}} = \frac{1}{2} \cdot \sqrt{1 - p^2} \] Finally, to summarize: - \( \cos 200^{\circ} = -p \) - \( \sin(-70^{\circ}) = -p \) - \( \sin 10^{\circ} = \sqrt{(1 - p^2)} \).