EXICRCISE 5 Calculate the following, without use of a calculator: \( \begin{array}{ll}\text { (a) } \cos 120^{\circ} & \text { (b) } \tan 315^{\circ} \\ \text { (c) } \sin ^{2} 240^{\circ} & \text { (d) } \frac{\sin 160^{\circ}}{\cos 250^{\circ}} \\ \text { (e) } \tan 225^{\circ}+\cos \left(-60^{\circ}\right)-\sin ^{2} 510^{\circ} & \text { (f) } \sin 168^{\circ}-\cos 78^{\circ}+\tan \left(-45^{\circ}\right) \\ \text { (g) } \frac{\tan 240^{\circ} \cdot \sin 115^{\circ}}{\cos 330^{\circ} \cdot \cos 205^{\circ}} & \text { (h) } \frac{\sin 210^{\circ} \cdot \cos 150^{\circ}}{\tan \left(-60^{\circ}\right) \cdot \tan 2330^{\circ}} \\ \text { (i) } \frac{\tan 225^{\circ} \cdot \sin 240^{\circ} \cdot \cos 330^{\circ}}{\cos \left(-210^{\circ}\right) \cdot \tan 150^{\circ}} & \text { (i) } \frac{\sin \left(-45^{\circ}\right) \cdot \cos 315^{\circ} \cdot \cos 215^{\circ}}{\sin 305^{\circ} \cdot \tan 750^{\circ} \cdot \tan \left(-300^{\circ}\right)} \\ \text { (k) } \frac{\sin 705^{\circ} \cdot \tan \left(-315^{\circ}\right) \cdot \cos 300^{\circ}}{\sin 150^{\circ} \cdot \cos \left(-75^{\circ}\right)} & \text { (l) } \frac{2 \sin 150^{\circ} \cos 325^{\circ}-\sin \left(-55^{\circ}\right)}{\cos 395^{\circ}} \\ \text { (m) } \frac{\tan \left(-60^{\circ}\right) \cdot \sin 158^{\circ}+\sin 120^{\circ} \cdot \cos 248^{\circ}}{\cos 570^{\circ} \cdot \cos 292^{\circ}} & \text { (n) } \frac{\sin 300^{\circ} \cos \left(-395^{\circ}\right)-\cos 210^{\circ} \sin 235^{\circ}}{\tan 150^{\circ} \sin 775^{\circ}+\cos \left(-30^{\circ}\right) \cos 215^{\circ}}\end{array} \)

To solve the trigonometric problems step by step, we will use known values and properties of trigonometric functions. Let's go through each part one by one. ### (a) \( \cos 120^{\circ} \) The angle \( 120^{\circ} \) is in the second quadrant where cosine is negative. We can find it using: \[ \cos 120^{\circ} = -\cos(180^{\circ} - 120^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2} \] ### (b) \( \tan 315^{\circ} \) The angle \( 315^{\circ} \) is in the fourth quadrant where tangent is negative. We can find it using: \[ \tan 315^{\circ} = -\tan(360^{\circ} - 315^{\circ}) = -\tan 45^{\circ} = -1 \] ### (c) \( \sin^{2} 240^{\circ} \) The angle \( 240^{\circ} \) is in the third quadrant where sine is negative. We can find it using: \[ \sin 240^{\circ} = -\sin(240^{\circ} - 180^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2} \] Thus, \[ \sin^{2} 240^{\circ} = \left(-\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4} \] ### (d) \( \frac{\sin 160^{\circ}}{\cos 250^{\circ}} \) Using the sine and cosine values: \[ \sin 160^{\circ} = \sin(180^{\circ} - 160^{\circ}) = \sin 20^{\circ} \] \[ \cos 250^{\circ} = -\cos(250^{\circ} - 180^{\circ}) = -\cos 70^{\circ = -\sin 20^{\circ} \] Thus, \[ \frac{\sin 160^{\circ}}{\cos 250^{\circ}} = \frac{\sin 20^{\circ}}{-\sin 20^{\circ}} = -1 \] ### (e) \( \tan 225^{\circ} + \cos(-60^{\circ}) - \sin^{2} 510^{\circ} \) Calculating each term: \[ \tan 225^{\circ} = \tan(180^{\circ} + 45^{\circ}) = \tan 45^{\circ} = 1 \] \[ \cos(-60^{\circ}) = \cos 60^{\circ} = \frac{1}{2} \] For \( \sin^{2} 510^{\circ} \): \[ 510^{\circ} = 510^{\circ} - 360^{\circ} = 150^{\circ} \quad \Rightarrow \quad \sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2} \] Thus, \[ \sin^{2} 510^{\circ} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4} \] Putting it all together: \[ 1 + \frac{1}{2} - \frac{1}{4} = 1 + 0.5 - 0.25 = 1.25 = \frac{5}{4} \] ### (f) \( \sin 168^{\circ} - \cos 78^{\circ} + \tan(-45^{\circ}) \) Calculating each term: \[ \sin 168^{\circ} = \sin(180^{\circ} - 12^{\circ}) = \sin 12^{\circ} \] \[ \cos 78^{\circ} = \sin 12^{\circ} \] \[ \tan(-45^{\circ}) = -1 \] Thus, \[ \sin 168^{\circ} - \cos 78^{\circ} + \tan(-45^{\circ}) = \sin 12^{\circ} - \sin 12^{\circ} - 1 = -1 \] ### (g) \( \frac{\tan 240^{\circ} \cdot \sin 115^{\circ}}{\cos 330^{\circ} \cdot \cos 205^{\circ}} \) Calculating each term: \[ \tan 240^{\circ} = \tan(180^{\circ} + 60^{\circ}) = \tan 60^{\circ} = \sqrt{3} \] \[ \sin 115^{\circ} = \sin(180^{\circ} - 65^{\circ}) = \sin 65^{\circ} \] \[ \cos 330^{\circ} = \cos(360^{\circ} - 30^{\circ}) = \cos 30^{\circ} = \frac{\sqrt{3}}{2} \] \[ \cos 205^{\circ} = -\cos(205^{\circ} - 180^{\circ}) = -\cos 25^{\circ} \] Thus, \[ \frac{\sqrt{3} \cdot \sin 115^{\circ}}{\frac{\sqrt{3}}{2} \cdot (-\cos 25^{\circ})} = \frac{2 \sin 115^{\circ}}{-\cos 25^{\circ}} = -2 \tan 115^{\circ} \] ### (h) \( \frac{\sin 210^{\circ} \cdot \cos 150^{\circ}}{\tan(-60^{\circ}) \cdot \tan 2330^{\circ}} \) Calculating each term: \[ \sin 210^{\circ} = -\frac{1}{2}, \quad \cos 150^{\circ} = -\frac{\sqrt{3}}{2} \] \[ \tan(-60^{\circ}) = -\sqrt{3}, \quad \tan 2330^{\circ} = \tan(2330^{\circ} - 2160^{\circ}) = \tan 170^{\circ} = -\tan 10^{\circ} \] Thus, \[ \frac{-\frac{1}{2} \cdot -\frac{\sqrt{3}}{2}}{-\sqrt{3} \cdot -\tan 10^{\circ}} = \frac{\frac{\sqrt{3}}{4}}