Answer :
We are given \(\sin(a) = -\frac{4}{5}\) and \(\cos(B) = -\frac{5}{8}\), with \(a\) in Quadrant III and \(B\) in Quadrant II. We want to find the exact value of \(\cos(a - B)\).
### Step 1: Find \(\cos(a)\)
Since \(a\) is in Quadrant III, both sine and cosine are negative in this quadrant. We use the Pythagorean identity to find \(\cos(a)\):
[tex]\[ \sin^2(a) + \cos^2(a) = 1 \][/tex]
Substitute \(\sin(a)\):
[tex]\[ \left(-\frac{4}{5}\right)^2 + \cos^2(a) = 1 \implies \frac{16}{25} + \cos^2(a) = 1 \][/tex]
Solving for \(\cos^2(a)\):
[tex]\[ \cos^2(a) = 1 - \frac{16}{25} = \frac{9}{25} \][/tex]
Since \(a\) is in Quadrant III (where cosine is negative), we have:
[tex]\[ \cos(a) = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \][/tex]
### Step 2: Find \(\sin(B)\)
Since \(B\) is in Quadrant II and sine is positive in this quadrant, we use the Pythagorean identity again to find \(\sin(B)\):
[tex]\[ \sin^2(B) + \cos^2(B) = 1 \][/tex]
Substitute \(\cos(B)\):
[tex]\[ \sin^2(B) + \left(-\frac{5}{8}\right)^2 = 1 \implies \sin^2(B) + \frac{25}{64} = 1 \][/tex]
Solving for \(\sin^2(B)\):
[tex]\[ \sin^2(B) = 1 - \frac{25}{64} = \frac{39}{64} \][/tex]
Since \(B\) is in Quadrant II (where sine is positive), we have:
[tex]\[ \sin(B) = \sqrt{\frac{39}{64}} = \frac{\sqrt{39}}{8} \][/tex]
### Step 3: Use the angle subtraction formula for cosine
The cosine angle reduction formula is:
[tex]\[ \cos(a - B) = \cos(a) \cos(B) + \sin(a) \sin(B) \][/tex]
Substitute the known values:
[tex]\[ \cos(a) = -\frac{3}{5}, \quad \cos(B) = -\frac{5}{8}, \quad \sin(a) = -\frac{4}{5}, \quad \sin(B) = \frac{\sqrt{39}}{8} \][/tex]
Calculate \(\cos(a - B)\):
[tex]\[ \cos(a - B) = \left(-\frac{3}{5}\right) \left(-\frac{5}{8}\right) + \left(-\frac{4}{5}\right) \left(\frac{\sqrt{39}}{8}\right) \][/tex]
[tex]\[ \cos(a - B) = \left(\frac{15}{40}\right) + \left(\frac{-4 \sqrt{39}}{40}\right) \][/tex]
[tex]\[ \cos(a - B) = \frac{15 - 4 \sqrt{39}}{40} \][/tex]
Therefore, the exact value of \(\cos(a - B)\) is:
[tex]\[ \boxed{\frac{15 - 4 \sqrt{39}}{40}} \][/tex]
### Step 1: Find \(\cos(a)\)
Since \(a\) is in Quadrant III, both sine and cosine are negative in this quadrant. We use the Pythagorean identity to find \(\cos(a)\):
[tex]\[ \sin^2(a) + \cos^2(a) = 1 \][/tex]
Substitute \(\sin(a)\):
[tex]\[ \left(-\frac{4}{5}\right)^2 + \cos^2(a) = 1 \implies \frac{16}{25} + \cos^2(a) = 1 \][/tex]
Solving for \(\cos^2(a)\):
[tex]\[ \cos^2(a) = 1 - \frac{16}{25} = \frac{9}{25} \][/tex]
Since \(a\) is in Quadrant III (where cosine is negative), we have:
[tex]\[ \cos(a) = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \][/tex]
### Step 2: Find \(\sin(B)\)
Since \(B\) is in Quadrant II and sine is positive in this quadrant, we use the Pythagorean identity again to find \(\sin(B)\):
[tex]\[ \sin^2(B) + \cos^2(B) = 1 \][/tex]
Substitute \(\cos(B)\):
[tex]\[ \sin^2(B) + \left(-\frac{5}{8}\right)^2 = 1 \implies \sin^2(B) + \frac{25}{64} = 1 \][/tex]
Solving for \(\sin^2(B)\):
[tex]\[ \sin^2(B) = 1 - \frac{25}{64} = \frac{39}{64} \][/tex]
Since \(B\) is in Quadrant II (where sine is positive), we have:
[tex]\[ \sin(B) = \sqrt{\frac{39}{64}} = \frac{\sqrt{39}}{8} \][/tex]
### Step 3: Use the angle subtraction formula for cosine
The cosine angle reduction formula is:
[tex]\[ \cos(a - B) = \cos(a) \cos(B) + \sin(a) \sin(B) \][/tex]
Substitute the known values:
[tex]\[ \cos(a) = -\frac{3}{5}, \quad \cos(B) = -\frac{5}{8}, \quad \sin(a) = -\frac{4}{5}, \quad \sin(B) = \frac{\sqrt{39}}{8} \][/tex]
Calculate \(\cos(a - B)\):
[tex]\[ \cos(a - B) = \left(-\frac{3}{5}\right) \left(-\frac{5}{8}\right) + \left(-\frac{4}{5}\right) \left(\frac{\sqrt{39}}{8}\right) \][/tex]
[tex]\[ \cos(a - B) = \left(\frac{15}{40}\right) + \left(\frac{-4 \sqrt{39}}{40}\right) \][/tex]
[tex]\[ \cos(a - B) = \frac{15 - 4 \sqrt{39}}{40} \][/tex]
Therefore, the exact value of \(\cos(a - B)\) is:
[tex]\[ \boxed{\frac{15 - 4 \sqrt{39}}{40}} \][/tex]
