EXERCISE 5 Calculate the following, without use of a calculator: \( \begin{array}{ll}\text { (a) } \cos 120^{\circ} & \text { (b) } \tan 315^{\circ} \\ \text { (c) } \sin ^{2} 240^{\circ} & \text { (d) } \frac{\sin 160^{\circ}}{\cos 250^{\circ}}\end{array} \)

To tackle these trigonometric problems, let’s break each one down step-by-step. (a) \( \cos 120^{\circ} \): 120 degrees lies in the second quadrant where the cosine value is negative. Applying the reference angle of \( 180^\circ - 120^\circ = 60^\circ \), we find \( \cos 120^{\circ} = -\cos 60^{\circ} = -\frac{1}{2} \). (b) \( \tan 315^{\circ} \): 315 degrees is in the fourth quadrant where tangent values are negative. The reference angle here is \( 360^\circ - 315^\circ = 45^\circ \), hence \( \tan 315^{\circ} = -\tan 45^{\circ} = -1 \). (c) \( \sin^2 240^{\circ} \): 240 degrees is in the third quadrant; its reference angle is \( 240^\circ - 180^\circ = 60^\circ \). Therefore, \( \sin 240^{\circ} = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2} \). Squaring this gives us \( \sin^2 240^{\circ} = \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \). (d) \( \frac{\sin 160^{\circ}}{\cos 250^{\circ}} \): For \( \sin 160^{\circ} \), since 160 degrees is in the second quadrant, its reference angle is \( 180^\circ - 160^\circ = 20^\circ \). So, \( \sin 160^{\circ} = \sin 20^{\circ} \). For \( \cos 250^{\circ} \), which is in the third quadrant, we find that its reference angle is \( 250^\circ - 180^\circ = 70^\circ \), thus \( \cos 250^{\circ} = -\cos 70^{\circ} \). This gives us: \[ \frac{\sin 20^{\circ}}{-\cos 70^{\circ}} = -\frac{\sin 20^{\circ}}{\sin 20^{\circ}} = -1 \quad \text{(since } \cos 70^{\circ} = \sin 20^{\circ} \text{)} \] To summarize: (a) \( \cos 120^{\circ} = -\frac{1}{2} \) (b) \( \tan 315^{\circ} = -1 \) (c) \( \sin^2 240^{\circ} = \frac{3}{4} \) (d) \( \frac{\sin 160^{\circ}}{\cos 250^{\circ}} = -1 \)