Trigonometry & Inverse Trigonometry
Trigonometric Identities
Grade 11
Question:
<p>If \(k_1 = \tan 27\theta - \tan 9\theta + \tan 9\theta - \tan 3\theta + \tan 3\theta - \tan \theta\) and \(k_2 = \frac{\sin 3\theta}{\cos 3\theta} + \frac{\sin 9\theta}{\cos 9\theta} + \frac{\sin 27\theta}{\cos 27\theta}\), then</p>
<p>(a) \(k_1 = k_2\)</p>
<p>(b) \(k_1 = 2k_2\)</p>
<p>(c) \(k_1 + k_2 = 2\)</p>
<p>(d) \(k_2 = 2k_1\)</p>
Step-by-Step Solution
Key Concept: Recognize the telescoping series in kâ and apply the tangent difference identity to relate it to kâ.
<p>We can write $k_1 = \tan 27\theta - \tan 9\theta + \tan 9\theta - \tan 3\theta + \tan 3\theta - \tan \theta$ (telescoping series)</p><p>Using the identity $\tan 3\alpha - \tan \alpha = \frac{\sin 2\alpha}{\cos 3\alpha \cos \alpha}$:</p><p>$k_1 = 2\left(\frac{\sin 3\theta}{\cos 3\theta} + \frac{\sin 9\theta}{\cos 9\theta} + \frac{\sin 27\theta}{\cos 27\theta}\right) = 2k_2$</p><p>Therefore, $k_1 = 2k_2$.</p>
Correct Answer: B