Trigonometric Identities and Equations - SS3 Mathematics Past Questions and Answers - page 1

\[\cos^{2}\theta + 2\cos\theta + 1 = 0\]

Let \(P = \cos\theta\)

\[P^{2} + 2P + 1 = 0\]

\[(P - 1)(P - 1) = 0\]

\[P - 1 = 0\]

\[P = 1\]

\[\cos\theta = 1\]

\[\theta = \cos^{- 1}1 = 0{^\circ}or\ 360{^\circ}\]

Thus, \(\theta = 0{^\circ}\ or\ 360{^\circ}\)

\[\tan{75{^\circ}} = \tan{(45{^\circ} + 30{^\circ})}\]

\[\tan{(A + B)} = \frac{\sin(A + B)}{\cos(A + B)} = \frac{\tan A + \tan B}{1 - \tan A\tan B}\]

\[\tan{(45{^\circ} + 30{^\circ})} = \frac{\sin(45{^\circ} + 30{^\circ})}{\cos(45{^\circ} + 30{^\circ})} = \frac{\tan{45{^\circ}} + \tan{30{^\circ}}}{1 - \tan{45{^\circ}}\tan{30{^\circ}}}\]

\[\tan{(45{^\circ} + 30{^\circ})} = \frac{\tan{45{^\circ}} + \tan{30{^\circ}}}{1 - \tan{45{^\circ}}\tan{30{^\circ}}} = \frac{1 + \frac{\sqrt{3}}{2}}{1 - (1)(\frac{\sqrt{3}}{2})} = \frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{\frac{2}{2} - \frac{\sqrt{3}}{2}}\]

\(\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}} = \frac{2 + \sqrt{3}}{2} \times \frac{2}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}\)

Since \(2\sin A\cos A = \sin{2A}\)

\[2\sin{2A}\cos{2A} = \sin{2(2A)}\]

\[= \sin{4A}\]

\[\cos(A + B) = \cos A\cos B - \sin A\sin B\]

\[\cos(120{^\circ} + 45{^\circ}) = \cos{120{^\circ}}\cos{45{^\circ}} - \sin{120{^\circ}}\sin{45{^\circ}}\]

Since \(\cos{120{^\circ}} = - \cos{60{^\circ}}\) and \(\sin{120{^\circ}} = \sin 60{^\circ}\)

\[\cos(120{^\circ} + 45{^\circ}) = - \cos{60{^\circ}}\cos{45{^\circ}} - \sin{60{^\circ}}\sin{45{^\circ}}\]

\(\cos(120{^\circ} + 45{^\circ}) = - \left( \frac{1}{2} \right)\left( \frac{\sqrt{2}}{2} \right) - \left( \frac{\sqrt{3}}{2} \right)\left( \frac{\sqrt{2}}{2} \right)\)

\(= - \frac{\sqrt{2}}{4} - \frac{\sqrt{2}.\sqrt{3}}{4} = \frac{- \sqrt{2} - \sqrt{2}.\sqrt{3}}{4} = \frac{- \sqrt{2}(1 + \sqrt{3})}{4}\)