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QUESTION 6 Sinplify the following expressions without using a calculator. \( 6.1 \quad \frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}} \) 6.2 \( [\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right] \) \( 6.3 \quad \) If \( \tan x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} \) and \( m^{2}+\frac{1}{m^{2}}=1 \) Calculate the value of \( x \) without the use of a calculator.

Ask by Munoz George. in South Africa
Jan 12,2025

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\( x = 120^{\circ} \) or \( x = 240^{\circ} \)

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Bonus Knowledge

Certainly! Let’s break down each expression step by step! For the first expression, we know: 1. **Sine and Cosine Values:** - \(\sin 210^{\circ} = -\frac{1}{2}\) - \(\cos 300^{\circ} = \frac{1}{2}\) - \(\tan 240^{\circ} = \frac{\sin 240^{\circ}}{\cos 240^{\circ}} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}\) - \(\cos 120^{\circ} = -\frac{1}{2}\) - \(\tan 150^{\circ} = -\frac{1}{\sqrt{3}}\) - \(\sin 330^{\circ} = -\frac{1}{2}\) Putting it all together: \[ \frac{\left(-\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\sqrt{3}\right)}{\left(-\frac{1}{2}\right) \left(-\frac{1}{\sqrt{3}}\right) \left(-\frac{1}{2}\right)} = \frac{-\frac{\sqrt{3}}{4}}{\frac{1}{4\sqrt{3}}} = -\sqrt{3} \cdot \frac{1}{3} = -1 \] --- For the second expression: \[ [\sin(-\theta) + \cos(360-\theta)][\cos(90-\theta) + \frac{\sin \theta}{\tan \theta}] \] Using identities: - \(\sin(-\theta) = -\sin \theta\) - \(\cos(360 - \theta) = \cos \theta\) - \(\cos(90 - \theta) = \sin \theta\) - \(\frac{\sin \theta}{\tan \theta} = \cos \theta\) Combining these: \[ [-\sin \theta + \cos \theta][\sin \theta + \cos \theta] = (\cos \theta - \sin \theta)(\sin \theta + \cos \theta) = \cos^2 \theta - \sin^2 \theta \] Using the identity \(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta\), we simplify to: \[ \cos(2\theta) \] --- For the third expression, if \(\tan x = m + \frac{1}{m}\) and \(m^2 + \frac{1}{m^2} = 1\), then knowing \(m^2 + \frac{1}{m^2} = (m + \frac{1}{m})^2 - 2\), we find that \(m + \frac{1}{m} = \pm 1\). Also, \(\tan x = m + \frac{1}{m}\) = 1, which implies \(x = 135^{\circ}\) or \(x = 315^{\circ}\), but since we are restricted to \(90^\circ \leq x \leq 270^\circ\), the answer is: \[ x = 135^{\circ} \] And that's it! You’ve simplified those expressions like a math wizard! 🧙‍♂️✨

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