Trigonometric Identities and Equations - SS3 Mathematics Past Questions and Answers - page 1
If \(x = \sin\theta\), simplify \(\frac{x}{\sqrt{1 - x^{2}}}\)
\(\cos^{2}\theta\)
\(\tan\theta\)
\(\sin\frac{\theta}{2}\)
\(\tan^{2}\theta\)
\(\frac{x}{\sqrt{1 - x^{2}}} = \ \frac{\sin\theta}{\sqrt{1 - {(\sin x)}^{2}}} = \frac{\sin\theta}{\sqrt{1 - \sin^{2}\theta}} = \frac{\sin\theta}{\sqrt{\cos^{2}\theta}} = \frac{\sin\theta}{\cos\theta} = \tan\theta\)
Find the value of \(\theta\) between \(0{^\circ}\) and \(360{^\circ}\) satisfying the equation \(\cos^{2}\theta + 2\cos\theta + 1 = 0\)
\(0{^\circ}\)
\(180{^\circ}\)
\(360{^\circ}\)
\(0{^\circ}\) and \(360{^\circ}\)
\[\cos^{2}\theta + 2\cos\theta + 1 = 0\]
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, \(\theta = 0{^\circ}\ or\ 360{^\circ}\)
Without using tables, find the value of \(\tan{75{^\circ}}\) in surd form
\(\frac{2 + \sqrt{3}}{2 - \sqrt{3}}\)
\(\frac{2 - \sqrt{3}}{2 - \sqrt{3}}\)
\(\frac{2}{2 + \sqrt{3}}\)
\(\frac{2}{\sqrt{3}}\)
\[\tan{75{^\circ}} = \tan{(45{^\circ} + 30{^\circ})}\]
\[\tan{(A + B)} = \frac{\sin(A + B)}{\cos(A + B)} = \frac{\tan A + \tan B}{1 - \tan A\tan B}\]
\[\tan{(45{^\circ} + 30{^\circ})} = \frac{\sin(45{^\circ} + 30{^\circ})}{\cos(45{^\circ} + 30{^\circ})} = \frac{\tan{45{^\circ}} + \tan{30{^\circ}}}{1 - \tan{45{^\circ}}\tan{30{^\circ}}}\]
\[\tan{(45{^\circ} + 30{^\circ})} = \frac{\tan{45{^\circ}} + \tan{30{^\circ}}}{1 - \tan{45{^\circ}}\tan{30{^\circ}}} = \frac{1 + \frac{\sqrt{3}}{2}}{1 - (1)(\frac{\sqrt{3}}{2})} = \frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{\frac{2}{2} - \frac{\sqrt{3}}{2}}\]
\(\frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}} = \frac{2 + \sqrt{3}}{2} \times \frac{2}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}\)
Simplify \(2\sin{2A}\cos{2A}\)
Since \(2\sin A\cos A = \sin{2A}\)
\[2\sin{2A}\cos{2A} = \sin{2(2A)}\]
\[= \sin{4A}\]
Without using tables, find the value of \(\cos{(120{^\circ} + 45{^\circ}})\) in surd form.
\[\cos(A + B) = \cos A\cos B - \sin A\sin B\]
\[\cos(120{^\circ} + 45{^\circ}) = \cos{120{^\circ}}\cos{45{^\circ}} - \sin{120{^\circ}}\sin{45{^\circ}}\]
Since \(\cos{120{^\circ}} = - \cos{60{^\circ}}\) and \(\sin{120{^\circ}} = \sin 60{^\circ}\)
\[\cos(120{^\circ} + 45{^\circ}) = - \cos{60{^\circ}}\cos{45{^\circ}} - \sin{60{^\circ}}\sin{45{^\circ}}\]
\(\cos(120{^\circ} + 45{^\circ}) = - \left( \frac{1}{2} \right)\left( \frac{\sqrt{2}}{2} \right) - \left( \frac{\sqrt{3}}{2} \right)\left( \frac{\sqrt{2}}{2} \right)\)
\(= - \frac{\sqrt{2}}{4} - \frac{\sqrt{2}.\sqrt{3}}{4} = \frac{- \sqrt{2} - \sqrt{2}.\sqrt{3}}{4} = \frac{- \sqrt{2}(1 + \sqrt{3})}{4}\)