Answer :
To find the value of \(\cos(A - B)\), we need to use the trigonometric identities and the given information about the angles.
Given:
- \(\sin (A) = \frac{4}{5}\) and \(\frac{\pi}{2} < A < \pi\)
- \(\sin (B) = \frac{-2 \sqrt{5}}{5}\) and \(\pi < B < \frac{3 \pi}{2}\)
Let's work through this step-by-step:
### Step 1: Determine \(\cos(A)\)
Since \(\frac{\pi}{2} < A < \pi\), angle \(A\) is in the second quadrant, where sine is positive and cosine is negative. We use the Pythagorean identity:
[tex]\[ \sin^2(A) + \cos^2(A) = 1 \][/tex]
Substituting the given value of \(\sin(A)\):
[tex]\[ \left(\frac{4}{5}\right)^2 + \cos^2(A) = 1 \][/tex]
[tex]\[ \frac{16}{25} + \cos^2(A) = 1 \][/tex]
[tex]\[ \cos^2(A) = 1 - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(A) = \frac{25}{25} - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(A) = \frac{9}{25} \][/tex]
[tex]\[ \cos(A) = \pm \frac{3}{5} \][/tex]
Since \(A\) is in the second quadrant, \(\cos(A)\) is negative:
[tex]\[ \cos(A) = -\frac{3}{5} \][/tex]
### Step 2: Determine \(\cos(B)\)
Since \(\pi < B < \frac{3 \pi}{2}\), angle \(B\) is in the third quadrant where both sine and cosine are negative. Again, we use the Pythagorean identity:
[tex]\[ \sin^2(B) + \cos^2(B) = 1 \][/tex]
Substituting the given value of \(\sin(B)\):
[tex]\[ \left(\frac{-2 \sqrt{5}}{5}\right)^2 + \cos^2(B) = 1 \][/tex]
[tex]\[ \frac{4 \cdot 5}{25} + \cos^2(B) = 1 \][/tex]
[tex]\[ \frac{20}{25} + \cos^2(B) = 1 \][/tex]
[tex]\[ \cos^2(B) = 1 - \frac{20}{25} \][/tex]
[tex]\[ \cos^2(B) = \frac{25}{25} - \frac{20}{25} \][/tex]
[tex]\[ \cos^2(B) = \frac{5}{25} \][/tex]
[tex]\[ \cos(B) = \pm \frac{\sqrt{5}}{5} \][/tex]
Since \(B\) is in the third quadrant, \(\cos(B)\) is negative:
[tex]\[ \cos(B) = -\frac{\sqrt{5}}{5} \][/tex]
### Step 3: Calculate \(\cos(A - B)\)
Now, we use the formula for \(\cos(A - B)\):
[tex]\[ \cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B) \][/tex]
Substitute the known values:
[tex]\[ \cos(A - B) = \left(-\frac{3}{5}\right)\left(-\frac{\sqrt{5}}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{-2 \sqrt{5}}{5}\right) \][/tex]
[tex]\[ \cos(A - B) = \frac{3 \sqrt{5}}{25} + \frac{-8 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = \frac{3 \sqrt{5} - 8 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = \frac{-5 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = -\frac{\sqrt{5}}{5} \][/tex]
Thus, the value of \(\cos(A - B)\) is \(-\frac{\sqrt{5}}{5}\).
So, the correct answer is:
[tex]\[ \boxed{-\frac{\sqrt{5}}{5}} \][/tex]
Given:
- \(\sin (A) = \frac{4}{5}\) and \(\frac{\pi}{2} < A < \pi\)
- \(\sin (B) = \frac{-2 \sqrt{5}}{5}\) and \(\pi < B < \frac{3 \pi}{2}\)
Let's work through this step-by-step:
### Step 1: Determine \(\cos(A)\)
Since \(\frac{\pi}{2} < A < \pi\), angle \(A\) is in the second quadrant, where sine is positive and cosine is negative. We use the Pythagorean identity:
[tex]\[ \sin^2(A) + \cos^2(A) = 1 \][/tex]
Substituting the given value of \(\sin(A)\):
[tex]\[ \left(\frac{4}{5}\right)^2 + \cos^2(A) = 1 \][/tex]
[tex]\[ \frac{16}{25} + \cos^2(A) = 1 \][/tex]
[tex]\[ \cos^2(A) = 1 - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(A) = \frac{25}{25} - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(A) = \frac{9}{25} \][/tex]
[tex]\[ \cos(A) = \pm \frac{3}{5} \][/tex]
Since \(A\) is in the second quadrant, \(\cos(A)\) is negative:
[tex]\[ \cos(A) = -\frac{3}{5} \][/tex]
### Step 2: Determine \(\cos(B)\)
Since \(\pi < B < \frac{3 \pi}{2}\), angle \(B\) is in the third quadrant where both sine and cosine are negative. Again, we use the Pythagorean identity:
[tex]\[ \sin^2(B) + \cos^2(B) = 1 \][/tex]
Substituting the given value of \(\sin(B)\):
[tex]\[ \left(\frac{-2 \sqrt{5}}{5}\right)^2 + \cos^2(B) = 1 \][/tex]
[tex]\[ \frac{4 \cdot 5}{25} + \cos^2(B) = 1 \][/tex]
[tex]\[ \frac{20}{25} + \cos^2(B) = 1 \][/tex]
[tex]\[ \cos^2(B) = 1 - \frac{20}{25} \][/tex]
[tex]\[ \cos^2(B) = \frac{25}{25} - \frac{20}{25} \][/tex]
[tex]\[ \cos^2(B) = \frac{5}{25} \][/tex]
[tex]\[ \cos(B) = \pm \frac{\sqrt{5}}{5} \][/tex]
Since \(B\) is in the third quadrant, \(\cos(B)\) is negative:
[tex]\[ \cos(B) = -\frac{\sqrt{5}}{5} \][/tex]
### Step 3: Calculate \(\cos(A - B)\)
Now, we use the formula for \(\cos(A - B)\):
[tex]\[ \cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B) \][/tex]
Substitute the known values:
[tex]\[ \cos(A - B) = \left(-\frac{3}{5}\right)\left(-\frac{\sqrt{5}}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{-2 \sqrt{5}}{5}\right) \][/tex]
[tex]\[ \cos(A - B) = \frac{3 \sqrt{5}}{25} + \frac{-8 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = \frac{3 \sqrt{5} - 8 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = \frac{-5 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = -\frac{\sqrt{5}}{5} \][/tex]
Thus, the value of \(\cos(A - B)\) is \(-\frac{\sqrt{5}}{5}\).
So, the correct answer is:
[tex]\[ \boxed{-\frac{\sqrt{5}}{5}} \][/tex]
