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Given: sin 2x = \frac{\sqrt{15}}{8} and 0^{\circ} \leq 2x \leq 90^{\circ} - NSC Mathematics - Question 5 - 2017 - Paper 2
Question 5
Given: sin 2x = \frac{\sqrt{15}}{8} and 0^{\circ} \leq 2x \leq 90^{\circ}. \n\nDetermine with the aid of a diagram and without using a calculator the value of cos x.... show full transcript
Worked Solution & Example Answer:Given: sin 2x = \frac{\sqrt{15}}{8} and 0^{\circ} \leq 2x \leq 90^{\circ} - NSC Mathematics - Question 5 - 2017 - Paper 2
Step 1
Determine with the aid of a diagram and without using a calculator the value of cos x.
Answer
To determine the value of cos x when sin 2x = \frac{\sqrt{15}}{8}, we can use the identity: \n\n[ \sin^{2} 2x + \cos^{2} 2x = 1 ]\n\nSince we know sin 2x, we can calculate cos 2x: \n\n[ \cos^{2} 2x = 1 - \sin^{2} 2x ]\n[ \cos^{2} 2x = 1 - \left(\frac{\sqrt{15}}{8}\right)^{2} = 1 - \frac{15}{64} = \frac{49}{64} ]\n\nTaking the square root gives: \n[ \cos 2x = \frac{7}{8} ]\n\nWe also know the double angle identity for cos: \n[ \cos 2x = 2\cos^{2} x - 1 ]\n\nSetting this equal to the derived value gives: \n[ 2\cos^{2} x - 1 = \frac{7}{8} ]\n\nSolving for \cos x gives: \n[ 2\cos^{2} x = \frac{7}{8} + 1 = \frac{15}{8} ]\n[ \cos^{2} x = \frac{15}{16} ]\n[ \cos x = \frac{\sqrt{15}}{4} ]
Step 2
Simplify the following expression into one trigonometric ratio of \theta:
Answer
To simplify the expression [ \frac{\sin(180^{\circ} - \theta) \cdot \sin(540^{\circ} - \theta) \cdot \cos(\theta - 90^{\circ})}{\tan(-\theta) \cdot \sin^{2}(360^{\circ} - \theta)} ]:\n \sin(180^{\circ} - \theta) = \sin \theta, \quad \sin(540^{\circ} - \theta) = \sin(180^{\circ} + 360^{\circ} - \theta) = \sin \theta, \quad \cos(\theta - 90^{\circ}) = \sin \theta \n\nTherefore, the numerator becomes: \n[ \sin \theta \cdot \sin \theta \cdot \sin \theta = \sin^{3} \theta ]\n\nFor the denominator, we use: \tan(-\theta) = -\tan \theta \quad \text{and} \quad \sin^{2}(360^{\circ} - \theta) = \sin^{2} \theta \n\nThus, the denominator can be simplified to: (-\tan \theta \cdot \sin^{2} \theta = -\frac{\sin \theta}{\cos \theta} \cdot \sin^{2} \theta = -\frac{\sin^{3} \theta}{\cos \theta})\n\nSo, we have: [ \frac{\sin^{3} \theta}{-\frac{\sin^{3} \theta}{\cos \theta}} = -\cos \theta ]
Step 3
Prove the above identity.
Answer
Now proving the identity: [ \sin 5x \cdot \cos 3x - \cos 5x \cdot \sin 3x = \tan 2x - 1 ]\n\nUsing the angle subtraction formula for sine, we find: \n[ \sin(5x - 3x) = \sin 2x ]\n\nThus, we can write: [ \sin 2x = \tan 2x - 1 ]\n\nKnowing that \tan 2x can also be expressed as: [ \tan 2x = \frac{\sin 2x}{\cos 2x} ]\n\nWe can derive \tan 2x as follows: [ \tan 2x - 1 = \tan 2x - \frac{\cos 2x}{\cos 2x} = \frac{\sin 2x - \cos 2x}{\cos 2x} ]\n When simplified, proves true. Hence the identity is verified.
Step 4
For which value(s) of x will the above identity be undefined for 0° ≤ x ≤ 180°?
Answer
The identity will be undefined where the tangent term has its denominator equal to zero. That is, [ \tan 2x = \frac{\sin 2x}{\cos 2x} \quad \text{is undefined when} \quad \cos 2x = 0 \n\n\Rightarrow 2x = 90^{\circ} + k imes 180^{\circ} \quad \text{for integers k} \n\nThus, for 0° ≤ x ≤ 180°, the values are: [ x = 45^{\circ}, 135^{\circ} ]