Step-by-step explanation:
To find the value of \( \sin(A-B) \), we can use the trigonometric identity:
\[ \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B) \]
Given that \( \cos(A) = \frac{20}{29} \) and \( \tan(B) = \frac{8}{15} \), we can use the definitions of sine, cosine, and tangent to find the missing trigonometric ratios.
First, let's find \( \sin(A) \):
\[ \sin^2(A) + \cos^2(A) = 1 \]
\[ \sin^2(A) = 1 - \cos^2(A) \]
\[ \sin(A) = \sqrt{1 - \cos^2(A)} \]
\[ \sin(A) = \sqrt{1 - \left(\frac{20}{29}\right)^2} \]
\[ \sin(A) = \frac{21}{29} \]
Next, let's find \( \cos(B) \) using the fact that \( \tan(B) = \frac{\sin(B)}{\cos(B)} \):
\[ \tan(B) = \frac{\sin(B)}{\cos(B)} \]
\[ \cos(B) = \frac{\sin(B)}{\tan(B)} \]
\[ \cos(B) = \frac{\frac{8}{15}}{\frac{8}{15}} \]
\[ \cos(B) = 1 \]
Now, let's find \( \sin(B) \) using the fact that \( \sin^2(B) + \cos^2(B) = 1 \):
\[ \sin^2(B) = 1 - \cos^2(B) \]
\[ \sin(B) = \sqrt{1 - \cos^2(B)} \]
\[ \sin(B) = \sqrt{1 - 1} \]
\[ \sin(B) = 0 \]
Now, we have all the necessary values to compute \( \sin(A - B) \):
\[ \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B) \]
\[ \sin(A - B) = \frac{21}{29} \times 1 - \frac{20}{29} \times 0 \]
\[ \sin(A - B) = \frac{21}{29} \]
So, the value of \( \sin(A - B) \) in simplest form is \( \frac{21}{29} \).