Pregunta
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Given: (i) \( A=60^{\circ} \) and \( B=30^{\circ} \) (ii) \( A=110^{\circ} \) and \( B=50^{\circ} \) (iii) \( A=225^{\circ} \) and \( B=135^{\circ} \) Use a calculator to evaluate each of the following: (a) \( \cos (A-B) \) (b) \( \cos A-\cos B \) (c) \( \cos A \cdot \cos B+\sin A \cdot \sin B \)

Ask by Dickson Marshall. in South Africa
Feb 04,2025

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Here are the results for each part: - **(a) \( \cos(A - B) \):** - \( A = 60^{\circ}, B = 30^{\circ} \): 0.8660 - \( A = 110^{\circ}, B = 50^{\circ} \): 0.5000 - \( A = 225^{\circ}, B = 135^{\circ} \): 0.0000 - **(b) \( \cos A - \cos B \):** - \( A = 60^{\circ}, B = 30^{\circ} \): -0.3660 - \( A = 110^{\circ}, B = 50^{\circ} \): -0.9848 - \( A = 225^{\circ}, B = 135^{\circ} \): 0.0000 - **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \):** - \( A = 60^{\circ}, B = 30^{\circ} \): 0.8660 - \( A = 110^{\circ}, B = 50^{\circ} \): 0.5000 - \( A = 225^{\circ}, B = 135^{\circ} \): 0.0000

Solución

Let's evaluate each part for the given angle pairs using a calculator. We'll work through each pair \((i)\), \((ii)\), and \((iii)\) separately. ### Given Angle Pairs: 1. **(i)** \( A = 60^{\circ} \), \( B = 30^{\circ} \) 2. **(ii)** \( A = 110^{\circ} \), \( B = 50^{\circ} \) 3. **(iii)** \( A = 225^{\circ} \), \( B = 135^{\circ} \) --- ### Part (a): \( \cos(A - B) \) 1. **(i)** \( \cos(60^{\circ} - 30^{\circ}) = \cos(30^{\circ}) \approx 0.8660 \) 2. **(ii)** \( \cos(110^{\circ} - 50^{\circ}) = \cos(60^{\circ}) = 0.5000 \) 3. **(iii)** \( \cos(225^{\circ} - 135^{\circ}) = \cos(90^{\circ}) = 0.0000 \) --- ### Part (b): \( \cos A - \cos B \) 1. **(i)** \[ \cos(60^{\circ}) - \cos(30^{\circ}) = 0.5000 - 0.8660 = -0.3660 \] 2. **(ii)** \[ \cos(110^{\circ}) - \cos(50^{\circ}) \approx (-0.3420) - 0.6428 = -0.9848 \] 3. **(iii)** \[ \cos(225^{\circ}) - \cos(135^{\circ}) \approx (-0.7071) - (-0.7071) = 0.0000 \] --- ### Part (c): \( \cos A \cdot \cos B + \sin A \cdot \sin B \) This expression is equivalent to \( \cos(A - B) \) based on the cosine difference identity. Therefore, the results will be the same as in part (a). 1. **(i)** \( \cos A \cdot \cos B + \sin A \cdot \sin B \approx 0.8660 \) 2. **(ii)** \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.5000 \) 3. **(iii)** \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.0000 \) --- ### Summary of Results: | Part | (i) \( A=60^{\circ}, B=30^{\circ} \) | (ii) \( A=110^{\circ}, B=50^{\circ} \) | (iii) \( A=225^{\circ}, B=135^{\circ} \) | |------|-------------------------------------|---------------------------------------|-----------------------------------------| | **a)** \( \cos(A - B) \) | 0.8660 | 0.5000 | 0.0000 | | **b)** \( \cos A - \cos B \) | -0.3660 | -0.9848 | 0.0000 | | **c)** \( \cos A \cos B + \sin A \sin B \) | 0.8660 | 0.5000 | 0.0000 | --- **Note:** All values are rounded to four decimal places.

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The Deep Dive

Let's evaluate each of the cases step by step! We'll use the fact that the cosine of the difference of two angles can be calculated using the formula \( \cos(A - B) = \cos A \cos B + \sin A \sin B \). ### Case (i) \( A=60^{\circ} \) and \( B=30^{\circ} \) 1. **(a) \( \cos (A - B) = \cos(60^{\circ} - 30^{\circ}) = \cos(30^{\circ}) \)** - This yields \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866 \) 2. **(b) \( \cos A - \cos B = \cos(60^{\circ}) - \cos(30^{\circ}) \)** - Calculation gives \( \frac{1}{2} - \frac{\sqrt{3}}{2} \approx -0.366 \) 3. **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)** - Results in \( \cos(60^{\circ}) \cdot \cos(30^{\circ}) + \sin(60^{\circ}) \cdot \sin(30^{\circ}) \) - Which calculates to \( \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \approx 0.866 \) ### Case (ii) \( A=110^{\circ} \) and \( B=50^{\circ} \) 1. **(a) \( \cos (A - B) = \cos(110^{\circ} - 50^{\circ}) = \cos(60^{\circ}) \)** - This yields \( \cos(60^{\circ}) = \frac{1}{2} \) 2. **(b) \( \cos A - \cos B = \cos(110^{\circ}) - \cos(50^{\circ}) \)** - Calculation gives approximately \( -0.342 - 0.643 \approx -0.985 \) 3. **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)** - Results in \( \cos(110^{\circ}) \cdot \cos(50^{\circ}) + \sin(110^{\circ}) \cdot \sin(50^{\circ}) \) - This will also give \( \frac{1}{2} \) ### Case (iii) \( A=225^{\circ} \) and \( B=135^{\circ} \) 1. **(a) \( \cos (A - B) = \cos(225^{\circ} - 135^{\circ}) = \cos(90^{\circ}) \)** - This yields \( \cos(90^{\circ}) = 0 \) 2. **(b) \( \cos A - \cos B = \cos(225^{\circ}) - \cos(135^{\circ}) \)** - Calculation gives \( -\frac{\sqrt{2}}{2} - (-\frac{\sqrt{2}}{2}) = 0 \) 3. **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)** - Results in \( (-\frac{\sqrt{2}}{2}) \cdot (-\frac{\sqrt{2}}{2}) + \sin(225^{\circ}) \cdot \sin(135^{\circ}) \) - This ultimately adds to \( 1 \) In conclusion, here are the results for each case: - Case (i) results: (0.866, -0.366, 0.866) - Case (ii) results: (0.5, -0.985, 0.5) - Case (iii) results: (0, 0, 1)

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