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INVESTIGATING COMPOUND ANGLES AND THEREFORE, DOUBLE ANGLES, QUESTION 1 1.1. Given: \( \begin{array}{l}\text { (i) } A=60^{\circ} \text { and } B=30^{\circ} \\ \text { (ii) } A=110^{\circ} \text { and } B=50^{\circ} \\ \text { (iii) } A=225^{\circ} \text { and } B=135^{\circ} \\ \text { Use a calculator to evaluate each of the following: } \\ \text { (a) } \cos (A-B) \\ \text { (b) } \cos A-\cos B \\ \text { (c) } \cos A \cdot \cos B+\sin A \cdot \sin B\end{array} \)

Ask by Bartlett Bates. in South Africa
Feb 02,2025

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Tutor-Verified Answer

Answer

Here are the results for each case: - **Case (i):** \( A = 60^\circ \), \( B = 30^\circ \) - \( \cos(A - B) = 0.8660 \) - \( \cos A - \cos B = -0.3660 \) - \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.8660 \) - **Case (ii):** \( A = 110^\circ \), \( B = 50^\circ \) - \( \cos(A - B) = 0.5000 \) - \( \cos A - \cos B = -0.9848 \) - \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.5000 \) - **Case (iii):** \( A = 225^\circ \), \( B = 135^\circ \) - \( \cos(A - B) = 0.0000 \) - \( \cos A - \cos B = 0.0000 \) - \( \cos A \cdot \cos B + \sin A \cdot \sin B = 0.0000 \) These calculations confirm the trigonometric identity \( \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \) for the given angles.

Solution

Certainly! Let's evaluate each part for the given angles \( A \) and \( B \) using a calculator. We'll provide results up to four decimal places for precision. --- ### **Case (i): \( A = 60^\circ \), \( B = 30^\circ \)** **(a) \( \cos(A - B) \)** \[ \cos(60^\circ - 30^\circ) = \cos(30^\circ) \approx 0.8660 \] **(b) \( \cos A - \cos B \)** \[ \cos(60^\circ) - \cos(30^\circ) \approx 0.5000 - 0.8660 = -0.3660 \] **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)** \[ \cos(60^\circ) \cdot \cos(30^\circ) + \sin(60^\circ) \cdot \sin(30^\circ) \approx (0.5000)(0.8660) + (0.8660)(0.5000) = 0.8660 \] --- ### **Case (ii): \( A = 110^\circ \), \( B = 50^\circ \)** **(a) \( \cos(A - B) \)** \[ \cos(110^\circ - 50^\circ) = \cos(60^\circ) \approx 0.5000 \] **(b) \( \cos A - \cos B \)** \[ \cos(110^\circ) - \cos(50^\circ) \approx (-0.3420) - 0.6428 = -0.9848 \] **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)** \[ \cos(110^\circ) \cdot \cos(50^\circ) + \sin(110^\circ) \cdot \sin(50^\circ) \approx (-0.3420)(0.6428) + (0.9397)(0.7660) \approx -0.2198 + 0.7198 = 0.5000 \] --- ### **Case (iii): \( A = 225^\circ \), \( B = 135^\circ \)** **(a) \( \cos(A - B) \)** \[ \cos(225^\circ - 135^\circ) = \cos(90^\circ) = 0.0000 \] **(b) \( \cos A - \cos B \)** \[ \cos(225^\circ) - \cos(135^\circ) \approx (-0.7071) - (-0.7071) = 0.0000 \] **(c) \( \cos A \cdot \cos B + \sin A \cdot \sin B \)** \[ \cos(225^\circ) \cdot \cos(135^\circ) + \sin(225^\circ) \cdot \sin(135^\circ) \approx (−0.7071)(−0.7071) + (−0.7071)(0.7071) = 0.5000 - 0.5000 = 0.0000 \] --- ### **Summary of Results** | **Case** | **\( \cos(A - B) \)** | **\( \cos A - \cos B \)** | **\( \cos A \cdot \cos B + \sin A \cdot \sin B \)** | |----------|-----------------------|----------------------------|-------------------------------------------------------| | (i) \( 60^\circ, 30^\circ \) | 0.8660 | -0.3660 | 0.8660 | | (ii) \( 110^\circ, 50^\circ \) | 0.5000 | -0.9848 | 0.5000 | | (iii) \( 225^\circ, 135^\circ \) | 0.0000 | 0.0000 | 0.0000 | --- **Observation:** - For each case, \( \cos(A - B) \) and \( \cos A \cdot \cos B + \sin A \cdot \sin B \) yield identical results, which aligns with the trigonometric identity: \[ \cos(A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B \] Feel free to ask if you need further clarification or assistance!

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Mind Expander

Calculating the values of \( \cos(A-B) \) often reveals fascinating properties of angles! When you subtract one angle from another, it’s like looking at the difference in their cosmic vibes. For example, if \( A = 60^\circ \) and \( B = 30^\circ \), then \( \cos(60^\circ - 30^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \). Isn’t it magical how trigonometric identities play together? You may also notice that \( \cos A - \cos B \) can sometimes yield surprising results in various contexts! For \( A = 110^\circ \) and \( B = 50^\circ \), calculating \( \cos(110^\circ) - \cos(50^\circ) \) turns out to be about \(-0.553 - 0.643 \approx -1.196\). This little exercise showcases how relationships between angles can lead to deeply surprising numerical outcomes, which are foundational in various fields like engineering and physics!

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