Question: Given: \(\sin(A) = \frac{4}{5}\), \(\frac{\pi}{2} < A < \pi\) and \(\sin(B) =…

Given: \(\sin(A) = \frac{4}{5}\), \(\frac{\pi}{2} < A < \pi\) and \(\sin(B) = \frac{-2\sqrt{5}}{5}\), \(\pi < B < \frac{3\pi}{2}\)

What is the value of \(\cos(A - B)\)?

Options:

1. \(-\frac{2\sqrt{5}}{25}\)
2. \(\frac{2\sqrt{5}}{5}\)
3. \(-\frac{\sqrt{5}}{5}\)
4. \(\frac{11\sqrt{5}}{25}\)

Solution

Given: \[ \sin(A) = \frac{4}{5}, \quad \frac{\pi}{2} < A < \pi \] \[ \sin(B) = -\frac{2\sqrt{3}}{5}, \quad \pi < B < \frac{3\pi}{2} \] We need to find \(\cos(A - B)\). First, use the identity for \(\cos(A - B)\): \[ \cos(A - B) = \cos A \cos B + \sin A \sin B \] Find \(\cos A\): Since \(\frac{\pi}{2} < A < \pi\), \(\cos A\) will be negative. \[ \sin^2 A + \cos^2 A = 1 \] \[ \left(\frac{4}{5}\right)^2 + \cos^2 A = 1 \] \[ \frac{16}{25} + \cos^2 A = 1 \] \[ \cos^2 A = 1 - \frac{16}{25} \] \[ \cos^2 A = \frac{9}{25} \] \[ \cos A = -\frac{3}{5} \] Find \(\cos B\): Since \(\pi < B < \frac{3\pi}{2}\), \(\cos B\) will be negative. \[ \sin^2 B + \cos^2 B = 1 \] \[ \left(-\frac{2\sqrt{3}}{5}\right)^2 + \cos^2 B = 1 \] \[ \frac{12}{25} + \cos^2 B = 1 \] \[ \cos^2 B = 1 - \frac{12}{25} \] \[ \cos^2 B = \frac{13}{25} \] \[ \cos B = -\frac{\sqrt{13}}{5} \] Substitute \(\cos A\), \(\cos B\), \(\sin A\), and \(\sin B\) into the formula for \(\cos(A - B)\): \[ \cos(A - B) = \left(-\frac{3}{5}\right)\left(-\frac{\sqrt{13}}{5}\right) + \left(\frac{4}{5}\right)\left(-\frac{2\sqrt{3}}{5}\right) \] \[ = \frac{3\sqrt{13}}{25} - \frac{8\sqrt{3}}{25} \] \[ = \frac{3\sqrt{13} - 8\sqrt{3}}{25} \] The value of \(\cos(A - B)\) is \(\frac{3\sqrt{13} - 8\sqrt{3}}{25}\).