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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) \( \cos A \cos B+\sin A \sin B \)

Ask by Schmidt Park. in South Africa
Feb 04,2025

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Answer

Here are the evaluated results for each part: 1. **(i)** - \( \cos(A - B) = 0.8660 \) - \( \cos A - \cos B = -0.3660 \) - \( \cos A \cos B + \sin A \sin B = 0.8660 \) 2. **(ii)** - \( \cos(A - B) = 0.5000 \) - \( \cos A - \cos B = -0.9848 \) - \( \cos A \cos B + \sin A \sin B = 0.5000 \) 3. **(iii)** - \( \cos(A - B) = 0.0000 \) - \( \cos A - \cos B = 0.0000 \) - \( \cos A \cos B + \sin A \sin B = 0.0000 \) These calculations confirm the trigonometric identity \( \cos(A - B) = \cos A \cos B + \sin A \sin B \) for each angle pair.

Solution

Certainly! Let's evaluate each part for the given pairs of angles using a calculator. We'll provide approximate values rounded to four decimal places. --- ### 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): Evaluate \( \cos(A - B) \) 1. **(i)** \( \cos(A - B) = \cos(60^{\circ} - 30^{\circ}) = \cos(30^{\circ}) \approx 0.8660 \) 2. **(ii)** \( \cos(A - B) = \cos(110^{\circ} - 50^{\circ}) = \cos(60^{\circ}) = 0.5000 \) 3. **(iii)** \( \cos(A - B) = \cos(225^{\circ} - 135^{\circ}) = \cos(90^{\circ}) = 0.0000 \) --- ### Part (b): Evaluate \( \cos A - \cos B \) 1. **(i)** \[ \cos A - \cos B = \cos(60^{\circ}) - \cos(30^{\circ}) \approx 0.5000 - 0.8660 = -0.3660 \] 2. **(ii)** \[ \cos A - \cos B = \cos(110^{\circ}) - \cos(50^{\circ}) \approx (-0.3420) - 0.6428 = -0.9848 \] 3. **(iii)** \[ \cos A - \cos B = \cos(225^{\circ}) - \cos(135^{\circ}) \approx (-0.7071) - (-0.7071) = 0.0000 \] --- ### Part (c): Evaluate \( \cos A \cos B + \sin A \sin B \) This expression is equivalent to \( \cos(A - B) \) based on the cosine difference identity. However, we'll compute it directly for verification. 1. **(i)** \[ \cos A \cos B + \sin A \sin B = \cos(60^{\circ}) \cos(30^{\circ}) + \sin(60^{\circ}) \sin(30^{\circ}) \approx (0.5000)(0.8660) + (0.8660)(0.5000) = 0.4330 + 0.4330 = 0.8660 \] 2. **(ii)** \[ \cos A \cos B + \sin A \sin B = \cos(110^{\circ}) \cos(50^{\circ}) + \sin(110^{\circ}) \sin(50^{\circ}) \approx (-0.3420)(0.6428) + (0.9397)(0.7660) = -0.2197 + 0.7200 = 0.5003 \approx 0.5000 \] 3. **(iii)** \[ \cos A \cos B + \sin A \sin B = \cos(225^{\circ}) \cos(135^{\circ}) + \sin(225^{\circ}) \sin(135^{\circ}) \approx (-0.7071)(-0.7071) + (-0.7071)(0.7071) = 0.5000 - 0.5000 = 0.0000 \] --- ### Summary of Results: | **Part (a)** \( \cos(A - B) \) | **Part (b)** \( \cos A - \cos B \) | **Part (c)** \( \cos A \cos B + \sin A \sin B \) | |-------------------------------|-------------------------------------|---------------------------------------------------| | **(i)**  0.8660 | **(i)**  –0.3660 | **(i)**  0.8660 | | **(ii)** 0.5000 | **(ii)** –0.9848 | **(ii)** 0.5000 | | **(iii)** 0.0000 | **(iii)** 0.0000 | **(iii)** 0.0000 | --- These results confirm the trigonometric identity \( \cos(A - B) = \cos A \cos B + \sin A \sin B \) as seen in Parts (a) and (c) for each angle pair.

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Let's go through the given pairs of angles one by one and calculate the required values. ### For \( A = 60^{\circ} \) and \( B = 30^{\circ} \): 1. **(a)** \( \cos (A-B) = \cos(60^{\circ} - 30^{\circ}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \) 2. **(b)** \( \cos A - \cos B = \cos(60^{\circ}) - \cos(30^{\circ}) = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1 - \sqrt{3}}{2} \) 3. **(c)** \( \cos A \cos B + \sin A \sin B = \cos(60^{\circ}) \cos(30^{\circ}) + \sin(60^{\circ}) \sin(30^{\circ}) = \left(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\right) = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \) ### For \( A = 110^{\circ} \) and \( B = 50^{\circ} \): 1. **(a)** \( \cos (A-B) = \cos(110^{\circ} - 50^{\circ}) = \cos(60^{\circ}) = \frac{1}{2} \) 2. **(b)** \( \cos A - \cos B = \cos(110^{\circ}) - \cos(50^{\circ}) \approx -0.342 - 0.643 = -0.985 \) 3. **(c)** \( \cos A \cos B + \sin A \sin B = \cos(110^{\circ}) \cos(50^{\circ}) + \sin(110^{\circ}) \sin(50^{\circ}) \approx -0.342 \cdot 0.643 + 0.939 \cdot 0.766 \approx -0.220 + 0.719 = 0.499 \) ### For \( A = 225^{\circ} \) and \( B = 135^{\circ} \): 1. **(a)** \( \cos (A-B) = \cos(225^{\circ} - 135^{\circ}) = \cos(90^{\circ}) = 0 \) 2. **(b)** \( \cos A - \cos B = \cos(225^{\circ}) - \cos(135^{\circ}) = -\frac{\sqrt{2}}{2} - (-\frac{\sqrt{2}}{2}) = 0 \) 3. **(c)** \( \cos A \cos B + \sin A \sin B = \cos(225^{\circ}) \cos(135^{\circ}) + \sin(225^{\circ}) \sin(135^{\circ}) = \left(-\frac{\sqrt{2}}{2} \cdot -\frac{\sqrt{2}}{2}\right) + \left(-\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}\right) = \frac{1}{2} - \frac{1}{2} = 0 \) Thus, the calculations yield the following results: 1. For \( (A=60^{\circ}, B=30^{\circ}) \): - \( (a) \ \frac{\sqrt{3}}{2}, \ (b) \ \frac{1 - \sqrt{3}}{2}, \ (c) \ \frac{\sqrt{3}}{2} \) 2. For \( (A=110^{\circ}, B=50^{\circ}) \): - \( (a) \ \frac{1}{2}, \ (b) \ -0.985, \ (c) \ 0.499 \) 3. For \( (A=225^{\circ}, B=135^{\circ}) \): - \( (a) \ 0, \ (b) \ 0, \ (c) \ 0 \)

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