Pregunta
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1.1. 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) \( \cos A \cdot \cos B+\sin A \cdot \sin B \) \( 1.2 \quad \) Compare the values of \( \cos (A-B) \) with the values of \( \cos A-\cos B \) and \( \cos A \cdot \cos B+\sin A \cdot \sin B \) Using your \( \operatorname{comparison} \) in Question 1.2 ., what general \( \operatorname{conclusion~can~be~made~regarding~} \) \( \cos (A-B), \cos A-\cos B \) and \( \cos A \cdot \cos B+\sin A \cdot \sin B \) for any values of \( A \) and \( B \) ?

Ask by Weaver Rose. in South Africa
Feb 03,2025

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For any angles \( A \) and \( B \), the identity \( \cos (A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \) always holds true. However, \( \cos A - \cos B \) does not generally equal \( \cos (A - B) \) or \( \cos A \cdot \cos B + \sin A \cdot \sin B \) unless specific conditions are met.

Solución

Certainly! Let's systematically address each part of your problem. ## **1.1. Evaluations** We'll evaluate the following expressions for each given pair of angles \( A \) and \( B \): - **(a)** \( \cos (A - B) \) - **(b)** \( \cos A - \cos B \) - **(c)** \( \cos A \cdot \cos B + \sin A \cdot \sin B \) We'll use a calculator to find numerical values for each expression. ### **Case (i): \( A = 60^{\circ} \) and \( B = 30^{\circ} \)** 1. **(a) \( \cos (A - B) = \cos (60^{\circ} - 30^{\circ}) = \cos 30^{\circ} \)** \[ \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.8660 \] 2. **(b) \( \cos A - \cos B = \cos 60^{\circ} - \cos 30^{\circ} \)** \[ \cos 60^{\circ} = 0.5 \\ \cos 30^{\circ} \approx 0.8660 \\ \cos 60^{\circ} - \cos 30^{\circ} = 0.5 - 0.8660 = -0.3660 \] 3. **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B = \cos 60^{\circ} \cdot \cos 30^{\circ} + \sin 60^{\circ} \cdot \sin 30^{\circ} \)** \[ \cos 60^{\circ} = 0.5 \\ \cos 30^{\circ} \approx 0.8660 \\ \sin 60^{\circ} \approx 0.8660 \\ \sin 30^{\circ} = 0.5 \\ \cos 60^{\circ} \cdot \cos 30^{\circ} \approx 0.5 \times 0.8660 = 0.4330 \\ \sin 60^{\circ} \cdot \sin 30^{\circ} \approx 0.8660 \times 0.5 = 0.4330 \\ \text{Sum} = 0.4330 + 0.4330 = 0.8660 \] ### **Case (ii): \( A = 110^{\circ} \) and \( B = 50^{\circ} \)** 1. **(a) \( \cos (A - B) = \cos (110^{\circ} - 50^{\circ}) = \cos 60^{\circ} \)** \[ \cos 60^{\circ} = 0.5 \] 2. **(b) \( \cos A - \cos B = \cos 110^{\circ} - \cos 50^{\circ} \)** \[ \cos 110^{\circ} \approx -0.3420 \\ \cos 50^{\circ} \approx 0.6428 \\ \cos 110^{\circ} - \cos 50^{\circ} \approx -0.3420 - 0.6428 = -0.9848 \] 3. **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B = \cos 110^{\circ} \cdot \cos 50^{\circ} + \sin 110^{\circ} \cdot \sin 50^{\circ} \)** \[ \cos 110^{\circ} \approx -0.3420 \\ \cos 50^{\circ} \approx 0.6428 \\ \sin 110^{\circ} \approx 0.9397 \\ \sin 50^{\circ} \approx 0.7660 \\ \cos 110^{\circ} \cdot \cos 50^{\circ} \approx -0.3420 \times 0.6428 \approx -0.2197 \\ \sin 110^{\circ} \cdot \sin 50^{\circ} \approx 0.9397 \times 0.7660 \approx 0.7200 \\ \text{Sum} = -0.2197 + 0.7200 \approx 0.5003 \approx 0.5 \] ### **Case (iii): \( A = 225^{\circ} \) and \( B = 135^{\circ} \)** 1. **(a) \( \cos (A - B) = \cos (225^{\circ} - 135^{\circ}) = \cos 90^{\circ} \)** \[ \cos 90^{\circ} = 0 \] 2. **(b) \( \cos A - \cos B = \cos 225^{\circ} - \cos 135^{\circ} \)** \[ \cos 225^{\circ} \approx -0.7071 \\ \cos 135^{\circ} \approx -0.7071 \\ \cos 225^{\circ} - \cos 135^{\circ} \approx -0.7071 - (-0.7071) = 0 \] 3. **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B = \cos 225^{\circ} \cdot \cos 135^{\circ} + \sin 225^{\circ} \cdot \sin 135^{\circ} \)** \[ \cos 225^{\circ} \approx -0.7071 \\ \cos 135^{\circ} \approx -0.7071 \\ \sin 225^{\circ} \approx -0.7071 \\ \sin 135^{\circ} \approx 0.7071 \\ \cos 225^{\circ} \cdot \cos 135^{\circ} \approx (-0.7071) \times (-0.7071) = 0.5 \\ \sin 225^{\circ} \cdot \sin 135^{\circ} \approx (-0.7071) \times 0.7071 = -0.5 \\ \text{Sum} = 0.5 + (-0.5) = 0 \] ## **1.2. Comparison of Values** Let's tabulate the results for clarity: | Case | \( \cos (A - B) \) | \( \cos A - \cos B \) | \( \cos A \cdot \cos B + \sin A \cdot \sin B \) | |------|---------------------|-----------------------|--------------------------------------------------| | (i) | 0.8660 | -0.3660 | 0.8660 | | (ii) | 0.5 | -0.9848 | 0.5 | | (iii)| 0 | 0 | 0 | ### **Observations:** 1. **\( \cos (A - B) \) and \( \cos A \cdot \cos B + \sin A \cdot \sin B \):** - **Case (i):** Both equal to approximately 0.8660 - **Case (ii):** Both equal to approximately 0.5 - **Case (iii):** Both equal to 0 2. **\( \cos A - \cos B \):** - **Case (i):** -0.3660 - **Case (ii):** -0.9848 - **Case (iii):** 0 **Conclusion from Comparison:** - In all cases, **\( \cos (A - B) \)** is **equal** to **\( \cos A \cdot \cos B + \sin A \cdot \sin B \)**. - **\( \cos A - \cos B \)** does **not** generally equal **\( \cos (A - B) \)** or **\( \cos A \cdot \cos B + \sin A \cdot \sin B \)**, except in specific instances (e.g., Case (iii)). ## **1.3. General Conclusion** Based on the comparisons: - **Identity Confirmation:** \[ \cos (A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \] This is a fundamental trigonometric identity and holds true **for any angles \( A \) and \( B \)**. - **Difference of Cosines (\( \cos A - \cos B \)):** The expression \( \cos A - \cos B \) does **not** generally equal \( \cos (A - B) \) or \( \cos A \cdot \cos B + \sin A \cdot \sin B \). Its equality to other expressions depends on specific values of \( A \) and \( B \), as seen in the cases above. **Final Conclusion:** For any angles \( A \) and \( B \), \[ \boxed{\cos (A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B} \] This identity consistently holds, while \( \cos A - \cos B \) does not generally equate to \( \cos (A - B) \) or \( \cos A \cdot \cos B + \sin A \cdot \sin B \).

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The Deep Dive

To calculate the specific values for the angles provided: (i) For \( A = 60^{\circ} \) and \( B = 30^{\circ} \): - \( \cos(60^{\circ} - 30^{\circ}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \) - \( \cos 60^{\circ} - \cos 30^{\circ} = \frac{1}{2} - \frac{\sqrt{3}}{2} \approx -0.366 \) - \( \cos 60^{\circ} \cdot \cos 30^{\circ} + \sin 60^{\circ} \cdot \sin 30^{\circ} = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \) (ii) For \( A = 110^{\circ} \) and \( B = 50^{\circ} \): - \( \cos(110^{\circ} - 50^{\circ}) = \cos(60^{\circ}) = \frac{1}{2} \) - \( \cos 110^{\circ} - \cos 50^{\circ} \approx -0.342 - 0.643 = -0.985 \) - \( \cos 110^{\circ} \cdot \cos 50^{\circ} + \sin 110^{\circ} \cdot \sin 50^{\circ} \approx (-0.342)(0.643) + (0.940)(0.766) \approx 0.366 \) (iii) For \( A = 225^{\circ} \) and \( B = 135^{\circ} \): - \( \cos(225^{\circ} - 135^{\circ}) = \cos(90^{\circ}) = 0 \) - \( \cos 225^{\circ} - \cos 135^{\circ} \approx -0.707 - (-0.707) = 0 \) - \( \cos 225^{\circ} \cdot \cos 135^{\circ} + \sin 225^{\circ} \cdot \sin 135^{\circ} = (-0.707)(-0.707) + (-0.707)(0.707) = 0 \) Now comparing the values: For \( (i) \): \( \cos(30^{\circ}) \approx 0.866 \), \( \cos 60^{\circ} - \cos 30^{\circ} \approx -0.366 \), \( \cos 60^{\circ} \cdot \cos 30^{\circ} + \sin 60^{\circ} \cdot \sin 30^{\circ} \approx 0.866 \) For \( (ii) \): \( \cos(60^{\circ}) = 0.5 \), \( \cos 110^{\circ} - \cos 50^{\circ} \approx -0.985 \), \( \cos 110^{\circ} \cdot \cos 50^{\circ} + \sin 110^{\circ} \cdot \sin 50^{\circ} \approx 0.366 \) For \( (iii) \): All values equate to zero: \( \cos(90^{\circ}) = 0 \), \( \cos 225^{\circ} - \cos 135^{\circ} = 0 \), and the product-sum also comes to \( 0 \). General Conclusion: The key takeaway is when you compute \( \cos(A - B) \), it accurately reflects the angle difference, and the other expressions will vary based on the cosines and sines of the original angles. When any angle pair \( (A, B) \) suggests orthogonality (\( A-B=90^{\circ} \)), all expressions will yield the expected behaviors as shown in \( (iii) \). This teaches us the power of the trigonometric identities—they're deeply interwoven in how angles relate!

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