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 the problem: \[ \sin(A) = \frac{4}{5}, \quad \frac{\pi}{2} < A < \pi \] \[ \sin(B) = -\frac{2\sqrt{5}}{5}, \quad \pi < B < \frac{3\pi}{2} \] We need to find the value of \(\cos(A - B)\). First, let’s find \(\cos(A)\): Since \(A\) is in the second quadrant, \(\cos(A)\) is negative. Using the Pythagorean identity: \[ \sin^2(A) + \cos^2(A) = 1 \] \[ \left(\frac{4}{5}\right)^2 + \cos^2(A) = 1 \] \[ \frac{16}{25} + \cos^2(A) = 1 \] \[ \cos^2(A) = 1 - \frac{16}{25} = \frac{9}{25} \] \[ \cos(A) = -\frac{3}{5} \] Now, let’s find \(\cos(B)\): Since \(B\) is in the third quadrant, \(\cos(B)\) is also negative. Again using the Pythagorean identity: \[ \sin^2(B) + \cos^2(B) = 1 \] \[ \left(-\frac{2\sqrt{5}}{5}\right)^2 + \cos^2(B) = 1 \] \[ \frac{20}{25} + \cos^2(B) = 1 \] \[ \cos^2(B) = 1 - \frac{20}{25} = \frac{5}{25} \] \[ \cos(B) = -\frac{\sqrt{5}}{5} \] Now, we use the cosine angle subtraction formula: \[ \cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B) \] \[ \cos(A - B) = \left(-\frac{3}{5}\right)\left(-\frac{\sqrt{5}}{5}\right) + \left(\frac{4}{5}\right)\left(-\frac{2\sqrt{5}}{5}\right) \] \[ \cos(A - B) = \frac{3\sqrt{5}}{25} - \frac{8\sqrt{5}}{25} \] \[ \cos(A - B) = \frac{3\sqrt{5} - 8\sqrt{5}}{25} \] \[ \cos(A - B) = \frac{-5\sqrt{5}}{25} \] \[ \cos(A - B) = -\frac{\sqrt{5}}{5} \] Thus, the value of \(\cos(A - B)\) is \(-\frac{\sqrt{5}}{5}\).