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QUESTION 6 6.1 Given: \( \tan x=\frac{3}{4} \), where \( x \in\left[180^{\circ} ; 270^{\circ}\right] \). With the aid of a sketch, and without the use of a calculator, calculate: 6.1.1 \( \sin x \) 6.1.2 \( \quad 2-\sin 2 x \) 6.1.3 \( \cos ^{2}\left(90^{\circ}-x\right)-1 \) 6.2 Evaluate: \( \frac{-1+\cos \left(180^{\circ}-\theta\right) \cdot \sin \left(\theta-90^{\circ}\right)}{\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right) \cdot \tan ^{2}\left(540^{\circ}+\theta\right)} \) QUESTION 7 7.1 Given: \( \cos (A-B)=\cos A \cos B+\sin A \sin B \) 7.1.1 Use the above identity to deduce that \( \sin (A+B)=\sin A \cos B+\cos \), 7.1.2 Hence determine the general solution of the equation \[ \sin \left(2 x+50^{\circ}\right)-\sin 15^{\circ} \cos 48^{\circ}=\sin 48^{\circ} \cos 15^{\circ} \] 7.2 Given: \( \frac{4 \sin x \cos x}{2 \sin ^{2} x-1} \) 7.2.1 Simplify \( \frac{4 \sin x \cos x}{2 \sin ^{2} x-1} \) to a single trigonometric ratio. 7.2.2 For which value(s) of \( x \) in the interval \( -90^{\circ}
Ask by Hodgson Barrett. in South Africa
Mar 12,2025

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**Simplified Answers:** *Question 6:* 1. \( \sin x = -\frac{3}{5} \) 2. \( 2 - \sin 2x = \frac{26}{25} \) 3. \( \cos^2(90^{\circ}-x) - 1 = -\frac{16}{25} \) 4. \( \frac{-1 + \cos(180^{\circ}-\theta) \cdot \sin(\theta-90^{\circ})}{\cos(-\theta) \cdot \sin(90^{\circ}+\theta) \cdot \tan^2(540^{\circ}+\theta)} = \frac{-1 + \cos^2 \theta}{\sin^2 \theta} \) *Question 7:* 1. \( \sin(A+B) = \sin A \cos B + \cos A \sin B \) 2. General solution: \( 2x + 50^{\circ} = 48^{\circ} + 15^{\circ} + n \cdot 360^{\circ} \) or \( 2x + 50^{\circ} = 180^{\circ} - (48^{\circ} + 15^{\circ}) + n \cdot 360^{\circ} \) 3. Simplified expression: \( -2 \tan(2x) \) 4. Undefined when \( x = 45^{\circ} + n \cdot 90^{\circ} \) 5. \( \frac{4 \sin 15^{\circ} \cos 15^{\circ}}{2 \sin^2 15^{\circ} - 1} = -2 \tan(30^{\circ}) = -2 \times \frac{\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3} \)

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Let's solve the problems step by step. ### QUESTION 6 #### 6.1 Given: \( \tan x=\frac{3}{4} \), where \( x \in\left[180^{\circ} ; 270^{\circ}\right] \). Since \( x \) is in the third quadrant, both sine and cosine will be negative. 1. **Finding \( \sin x \)**: - We know that \( \tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} \). - Using the Pythagorean theorem, we can find the hypotenuse \( r \): \[ r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] - Therefore, the coordinates are \( (-4, -3) \) in the third quadrant. - Thus, \( \sin x = \frac{\text{opposite}}{r} = \frac{-3}{5} \). 2. **Finding \( 2 - \sin 2x \)**: - We use the double angle formula: \( \sin 2x = 2 \sin x \cos x \). - First, we need \( \cos x \): \[ \cos x = \frac{\text{adjacent}}{r} = \frac{-4}{5} \] - Now calculate \( \sin 2x \): \[ \sin 2x = 2 \cdot \left(-\frac{3}{5}\right) \cdot \left(-\frac{4}{5}\right) = 2 \cdot \frac{12}{25} = \frac{24}{25} \] - Therefore, \( 2 - \sin 2x = 2 - \frac{24}{25} = \frac{50}{25} - \frac{24}{25} = \frac{26}{25} \). 3. **Finding \( \cos^2(90^{\circ}-x) - 1 \)**: - We know that \( \cos(90^{\circ}-x) = \sin x \). - Therefore, \( \cos^2(90^{\circ}-x) = \sin^2 x = \left(-\frac{3}{5}\right)^2 = \frac{9}{25} \). - Thus, \( \cos^2(90^{\circ}-x) - 1 = \frac{9}{25} - 1 = \frac{9}{25} - \frac{25}{25} = -\frac{16}{25} \). #### 6.2 Evaluate: \[ \frac{-1+\cos(180^{\circ}-\theta) \cdot \sin(\theta-90^{\circ})}{\cos(-\theta) \cdot \sin(90^{\circ}+\theta) \cdot \tan^2(540^{\circ}+\theta)} \] - We know: - \( \cos(180^{\circ}-\theta) = -\cos \theta \) - \( \sin(\theta-90^{\circ}) = -\cos \theta \) - \( \cos(-\theta) = \cos \theta \) - \( \sin(90^{\circ}+\theta) = \cos \theta \) - \( \tan(540^{\circ}+\theta) = \tan(\theta) \) Now substituting these values: \[ \frac{-1 + (-\cos \theta)(-\cos \theta)}{\cos \theta \cdot \cos \theta \cdot \tan^2(\theta)} = \frac{-1 + \cos^2 \theta}{\cos^2 \theta \cdot \tan^2 \theta} \] Since \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \): \[ = \frac{-1 + \cos^2 \theta}{\cos^2 \theta \cdot \frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{-1 + \cos^2 \theta}{\sin^2 \theta} \] ### QUESTION 7 #### 7.1 Given: \( \cos(A-B) = \cos A \cos B + \sin A \sin B \) 1. **Deducing \( \sin(A+B) \)**: - We know that \( \sin(A+B) = \sin A \cos B + \cos A \sin B \). - This can be derived from the identity for \( \sin(A+B) \) using the sine and cosine definitions. 2. **General solution of the equation**: \[ \sin(2x + 50^{\circ}) - \sin 15^{\circ} \cos 48^{\circ} = \sin 48^{\circ} \cos 15^{\circ} \] - Using the sine subtraction formula: \[ \sin(2x + 50^{\circ}) = \sin(48^{\circ} + 15^{\circ}) \] - Thus, \( 2x + 50^{\circ} = 48^{\circ} + 15^{\circ} + n \cdot 360^{\circ} \) or \( 2x + 50^{\circ} = 180^{\circ} - (48^{\circ} + 15^{\circ}) + n \cdot 360^{\circ} \). #### 7.2 Given: \( \frac{4 \sin x \cos x}{2 \sin^2 x - 1} \) 1. **Simplifying**: - We know \( 4 \sin x \cos x = 2 \sin(2x) \). - The denominator \( 2 \sin^2 x - 1 = -\cos(2x) \). - Thus, the expression simplifies to: \[ \frac{2 \sin(2x)}{-\cos(2x)} = -2 \tan(2x) \] 2. **Undefined values**: - The expression is undefined when \( 2 \sin^2 x - 1 = 0 \) or \( \cos(2x) = 0 \). - This occurs when \( 2x = 90^{\circ} + n \cdot 180^{\circ} \) or \( x = 45^{\circ} + n \cdot 90^{\circ} \). 3. **Calculating \( \frac{4 \sin 15^{\circ} \cos 15^{\circ}}{2 \sin^2 15^{\circ}

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The study of trigonometric functions dates back to ancient civilizations, including the Egyptians and Greeks, who used them for astronomy and architecture. The famous mathematician Hipparchus is often credited with creating the first known trigonometric table, providing the foundation for the trigonometric identities we still rely on today. Understanding these functions is key not just for solving angles, but also for real-world applications ranging from engineering to physics! When working with trigonometric functions, common mistakes include confusing the signs for different quadrants and misapplying identities. For instance, remember that in the third quadrant, both sine and cosine are negative, affecting your calculations of angles. Always double-check your quadrant placement and the values of each function – practice sketching triangles to visualize where the angles fall and what the signs should be!

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