If A is in the 3rd quardrant and B is the the fourth quadrant and cos A =`-15/17` , cos B =`4/5`, find the value of cos (A+B).

Answer

To find the value of \( \cos(A + B) \) given that \( A \) is in the 3rd quadrant with \( \cos A = -\frac{15}{17} \) and \( B \) is in the 4th quadrant with \( \cos B = \frac{4}{5} \), we will use the cosine addition formula:

\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]

### Step 1: Find \( \sin A \)
Since \( \cos A = -\frac{15}{17} \), we can find \( \sin A \) using the Pythagorean identity:

\[ \sin^2 A + \cos^2 A = 1 \]

Substituting the value of \( \cos A \):

\[ \sin^2 A + \left(-\frac{15}{17}\right)^2 = 1 \]

Calculating \( \left(-\frac{15}{17}\right)^2 \):

\[ \sin^2 A + \frac{225}{289} = 1 \]

Subtracting \( \frac{225}{289} \) from both sides:

\[ \sin^2 A = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289} \]

Taking the square root:

\[ \sin A = -\sqrt{\frac{64}{289}} = -\frac{8}{17} \]

(Note: We take the negative root because \( A \) is in the 3rd quadrant where sine is negative.)

### Step 2: Find \( \sin B \)
Since \( \cos B = \frac{4}{5} \), we can find \( \sin B \):

\[ \sin^2 B + \cos^2 B = 1 \]

Substituting the value of \( \cos B \):

\[ \sin^2 B + \left(\frac{4}{5}\right)^2 = 1 \]

Calculating \( \left(\frac{4}{5}\right)^2 \):

\[ \sin^2 B + \frac{16}{25} = 1 \]

Subtracting \( \frac{16}{25} \) from both sides:

\[ \sin^2 B = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25} \]

Taking the square root:

\[ \sin B = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \]

(Note: We take the negative root because \( B \) is in the 4th quadrant where sine is negative.)

### Step 3: Substitute values into the cosine addition formula
Now we substitute \( \cos A \), \( \cos B \), \( \sin A \), and \( \sin B \) into the cosine addition formula:

\[ \cos(A + B) = \left(-\frac{15}{17}\right) \left(\frac{4}{5}\right) - \left(-\frac{8}{17}\right) \left(-\frac{3}{5}\right) \]

Calculating each term:

1. \( \cos A \cos B = -\frac{15 \cdot 4}{17 \cdot 5} = -\frac{60}{85} \)
2. \( \sin A \sin B = -\frac{8 \cdot 3}{17 \cdot 5} = -\frac{24}{85} \)

Now substituting back:

\[ \cos(A + B) = -\frac{60}{85} - \frac{24}{85} \]

Combining the fractions:

\[ \cos(A + B) = -\frac{60 + 24}{85} = -\frac{84}{85} \]

### Final Answer
Thus, the value of \( \cos(A + B) \) is:

\[ \cos(A + B) = -\frac{84}{85} \]