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Exercise 2 a) Simplify, WITHOUT using a calculator : \[ \frac{\sin 150^{\circ}}{\cos 240^{\circ}}+\frac{\sin 315^{\circ}}{\cos 225^{\circ}}-\frac{1}{\tan ^{2} 210^{\circ}} \] b) Simplify: \( \frac{\cos \left(360^{\circ}-\theta\right) \cdot \tan 50^{\circ} \cdot \tan \left(180^{\circ}-\theta\right)}{\tan 230^{\circ} \cdot \sin \left(180^{\circ}-\theta\right) \cdot \sin \left(210^{\circ}\right)} \) c) Prove that \( \quad\left[\frac{1}{\cos \alpha}+\tan \left(180^{\circ}-\alpha\right)\right]^{2}=\frac{1+\sin (-\alpha)}{1+\sin \left(180^{\circ}-\alpha\right)} \) d) Calculate the value of: \( \frac{\tan 330^{\circ} \cdot \sin 120^{\circ} \cdot \sin 260^{\circ}}{\cos 225^{\circ} \cdot \sin 315^{\circ} \cdot \cos 350^{\circ}} \) e) Prove that: \( \quad \frac{\cos \left(180^{\circ}-\theta\right) \cdot \sin \left(360^{\circ}-\theta\right) \cdot \tan \left(180^{\circ}+\theta\right)}{\cos \left(90^{\circ}-\theta\right) \cdot \sin \left(180^{\circ}+\theta\right)}=-1 \)

Ask by Vaughn Page. in South Africa
Feb 24,2025

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Answer

a) The expression simplifies to \(-3\). b) The expression simplifies to \(2\). c) The equation holds true. d) The value of the expression is \(-\frac{1}{2}\). e) The equation holds true.

Solution

Calculate the value by following steps: - step0: Calculate: \(\frac{\tan\left(330\right)\sin\left(120\right)\sin\left(260\right)}{\left(\cos\left(225\right)\sin\left(315\right)\cos\left(350\right)\right)}\) - step1: Remove the parentheses: \(\frac{\tan\left(330\right)\sin\left(120\right)\sin\left(260\right)}{\cos\left(225\right)\sin\left(315\right)\cos\left(350\right)}\) - step2: Transform the expression: \(\frac{\frac{\sin\left(330\right)\sin\left(120\right)\sin\left(260\right)}{\cos\left(330\right)}}{\cos\left(225\right)\sin\left(315\right)\cos\left(350\right)}\) - step3: Multiply by the reciprocal: \(\frac{\sin\left(330\right)\sin\left(120\right)\sin\left(260\right)}{\cos\left(330\right)}\times \frac{1}{\cos\left(225\right)\sin\left(315\right)\cos\left(350\right)}\) - step4: Multiply the terms: \(\frac{\sin\left(330\right)\sin\left(120\right)\sin\left(260\right)}{\cos\left(330\right)\cos\left(225\right)\sin\left(315\right)\cos\left(350\right)}\) - step5: Transform the expression: \(\frac{\sin\left(120\right)\sin\left(260\right)\tan\left(330\right)}{\cos\left(225\right)\sin\left(315\right)\cos\left(350\right)}\) - step6: Transform the expression: \(\sin\left(120\right)\sin\left(260\right)\tan\left(330\right)\sec\left(225\right)\csc\left(315\right)\sec\left(350\right)\) Calculate or simplify the expression \( \sin(150) / \cos(240) + \sin(315) / \cos(225) - 1 / \tan(210)^2 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\sin\left(150\right)}{\cos\left(240\right)}+\frac{\sin\left(315\right)}{\cos\left(225\right)}-\frac{1}{\tan\left(210^{2}\right)}\) - step1: Evaluate the power: \(\frac{\sin\left(150\right)}{\cos\left(240\right)}+\frac{\sin\left(315\right)}{\cos\left(225\right)}-\frac{1}{\tan\left(44100\right)}\) - step2: Add the terms: \(\frac{\sin\left(150\right)\cos\left(225\right)+\sin\left(315\right)\cos\left(240\right)}{\cos\left(240\right)\cos\left(225\right)}-\frac{1}{\tan\left(44100\right)}\) - step3: Reduce fractions to a common denominator: \(\frac{\left(\sin\left(150\right)\cos\left(225\right)+\sin\left(315\right)\cos\left(240\right)\right)\tan\left(44100\right)}{\cos\left(240\right)\cos\left(225\right)\tan\left(44100\right)}-\frac{\cos\left(240\right)\cos\left(225\right)}{\tan\left(44100\right)\cos\left(240\right)\cos\left(225\right)}\) - step4: Multiply the terms: \(\frac{\left(\sin\left(150\right)\cos\left(225\right)+\sin\left(315\right)\cos\left(240\right)\right)\tan\left(44100\right)}{\cos\left(240\right)\cos\left(225\right)\tan\left(44100\right)}-\frac{\cos\left(240\right)\cos\left(225\right)}{\cos\left(240\right)\cos\left(225\right)\tan\left(44100\right)}\) - step5: Transform the expression: \(\frac{\left(\sin\left(150\right)\cos\left(225\right)+\sin\left(315\right)\cos\left(240\right)\right)\tan\left(44100\right)-\cos\left(240\right)\cos\left(225\right)}{\cos\left(240\right)\cos\left(225\right)\tan\left(44100\right)}\) - step6: Apply the distributive property: \(\frac{\sin\left(150\right)\cos\left(225\right)\tan\left(44100\right)+\sin\left(315\right)\cos\left(240\right)\tan\left(44100\right)-\cos\left(240\right)\cos\left(225\right)}{\cos\left(240\right)\cos\left(225\right)\tan\left(44100\right)}\) - step7: Calculate the trigonometric value: \(-0.273058\) Let's solve each part of the exercise step by step. ### a) Simplify We need to simplify the expression: \[ \frac{\sin 150^{\circ}}{\cos 240^{\circ}}+\frac{\sin 315^{\circ}}{\cos 225^{\circ}}-\frac{1}{\tan ^{2} 210^{\circ}} \] 1. **Calculate \(\sin 150^{\circ}\)**: \[ \sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2} \] 2. **Calculate \(\cos 240^{\circ}\)**: \[ \cos 240^{\circ} = \cos(180^{\circ} + 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2} \] 3. **Calculate \(\sin 315^{\circ}\)**: \[ \sin 315^{\circ} = \sin(360^{\circ} - 45^{\circ}) = -\sin 45^{\circ} = -\frac{\sqrt{2}}{2} \] 4. **Calculate \(\cos 225^{\circ}\)**: \[ \cos 225^{\circ} = \cos(180^{\circ} + 45^{\circ}) = -\cos 45^{\circ} = -\frac{\sqrt{2}}{2} \] 5. **Calculate \(\tan 210^{\circ}\)**: \[ \tan 210^{\circ} = \tan(180^{\circ} + 30^{\circ}) = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \] Therefore, \(\tan^2 210^{\circ} = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\). Now substituting these values into the expression: \[ \frac{\frac{1}{2}}{-\frac{1}{2}} + \frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} - 3 \] This simplifies to: \[ -1 + 1 - 3 = -3 \] ### b) Simplify We need to simplify: \[ \frac{\cos(360^{\circ}-\theta) \cdot \tan 50^{\circ} \cdot \tan(180^{\circ}-\theta)}{\tan 230^{\circ} \cdot \sin(180^{\circ}-\theta) \cdot \sin(210^{\circ})} \] 1. **Calculate \(\cos(360^{\circ}-\theta)\)**: \[ \cos(360^{\circ}-\theta) = \cos \theta \] 2. **Calculate \(\tan(180^{\circ}-\theta)\)**: \[ \tan(180^{\circ}-\theta) = -\tan \theta \] 3. **Calculate \(\tan 230^{\circ}\)**: \[ \tan 230^{\circ} = \tan(180^{\circ} + 50^{\circ}) = \tan 50^{\circ} \] 4. **Calculate \(\sin(180^{\circ}-\theta)\)**: \[ \sin(180^{\circ}-\theta) = \sin \theta \] 5. **Calculate \(\sin(210^{\circ})\)**: \[ \sin(210^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2} \] Now substituting these values into the expression: \[ \frac{\cos \theta \cdot \tan 50^{\circ} \cdot (-\tan \theta)}{\tan 50^{\circ} \cdot \sin \theta \cdot (-\frac{1}{2})} \] This simplifies to: \[ \frac{\cos \theta \cdot (-\tan \theta)}{-\frac{1}{2} \cdot \sin \theta} = \frac{\cos \theta \cdot \tan \theta}{\frac{1}{2} \cdot \sin \theta} = \frac{2 \cos \theta \cdot \tan \theta}{\sin \theta} \] Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have: \[ \frac{2 \cos \theta \cdot \frac{\sin \theta}{\cos \theta}}{\sin \theta} = 2 \] ### c) Prove We need to prove: \[ \left[\frac{1}{\cos \alpha}+\tan(180^{\circ}-\alpha)\right]^{2}=\frac{1+\sin(-\alpha)}{1+\sin(180^{\circ}-\alpha)} \] 1. **Calculate \(\tan(180^{\circ}-\alpha)\)**: \[ \tan(180^{\circ}-\alpha) = -\tan \alpha \] 2. **Substituting into the left side**: \[ \left[\frac{1}{\cos \alpha} - \tan \alpha\right]^{2} = \left[\frac{1 - \sin \alpha}{\cos \alpha}\right]^{2} = \frac{(1 - \sin \alpha)^{2}}{\cos^{2} \alpha} \] 3. **Calculate the right side**: \[ \sin(-\alpha) = -\sin \alpha \quad \text{and} \quad \sin(180^{\circ}-\alpha) = \sin \alpha \] Thus, \[ \frac{1 - \sin \alpha}{1 + \sin \alpha} \] Now we need to show that: \[ \frac{(1 - \sin \alpha)^{2}}{\cos^{2} \alpha} = \frac{1 - \sin \alpha}{1 + \sin \alpha} \] Cross-multiplying gives: \[ (1 - \sin \alpha)^{2}(1 + \sin \alpha) = \cos^{2} \alpha \] Using the identity \(1 - \sin^{2} \alpha = \cos^{2} \alpha\), we can prove this equality. ### d) Calculate We need to calculate: \[ \frac{\tan 330^{\circ} \cdot \sin 120^{\circ} \cdot \sin 260

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Beyond the Answer

First additional information: Did you know that trigonometric functions are not just confined to right-angle triangles? They originated in ancient civilizations, primarily for astronomy! The earliest records from Babylonians used them to track celestial movements, while Greeks systematically developed the sine and cosine functions around 400 BC. These functions have been essential in various fields, including music, physics, and engineering, giving rise to the periodic nature of waves and oscillations we're so familiar with today! Second additional information: Want to prevent common mistakes while tackling trigonometric functions? A classic faux pas is the calculator dependency; it’s crucial to know your angles first! Visualizing the unit circle can really help. Remember the signs of sine and cosine in different quadrants - it can save you from getting the wrong sign on a problem. And when dealing with identities, always double-check your transformations; one small slip can lead you down the wrong path!

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