To derive the relationship \(\Gamma(n)\Gamma(n-1) = \frac{\pi}{\sin(n\pi)}\), we will utilize properties of the Gamma function and Euler's reflection formula. Let's go through the steps.

1. **Gamma Function Definition:**

   The Gamma function \(\Gamma(n)\) is defined as:
   \[
   \Gamma(n) = \int_0^\infty t^{n-1} e^{-t} \, dt
   \]
   and it generalizes the factorial function, such that \(\Gamma(n) = (n-1)!\) for a positive integer \(n\).

2. **Euler's Reflection Formula:**

   Euler's reflection formula for the Gamma function states:
   \[
   \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}
   \]
   for any complex number \(z\).

3. **Applying Euler's Reflection Formula:**

   To derive \(\Gamma(n)\Gamma(n-1) = \frac{\pi}{\sin(n\pi)}\), we need to appropriately use Euler's reflection formula. Let's rewrite the terms in a suitable form.

   Consider the term \(\Gamma(n)\). By the reflection formula, if we take \(z = n\), we get:
   \[
   \Gamma(n)\Gamma(1-n) = \frac{\pi}{\sin(\pi n)}
   \]

   Notice that we need \(\Gamma(n-1)\) in our equation, but the reflection formula gives us \(\Gamma(1-n)\). To bridge this gap, use the recurrence relation of the Gamma function:
   \[
   \Gamma(n) = (n-1)\Gamma(n-1)
   \]

4. **Using the Recurrence Relation:**

   Substitute \(\Gamma(n-1)\) from the recurrence relation into the equation derived from the reflection formula:
   \[
   \Gamma(n) = (n-1)\Gamma(n-1)
   \]

   Therefore, we can express \(\Gamma(n-1)\) as:
   \[
   \Gamma(n-1) = \frac{\Gamma(n)}{n-1}
   \]

5. **Combining Both Results:**

   Now substitute \(\Gamma(1-n)\) back into Euler's reflection formula result:
   \[
   \Gamma(n)\Gamma(1-n) = \frac{\pi}{\sin(\pi n)}
   \]
   and use the fact that \(\Gamma(1-n) = \frac{\pi}{\sin(\pi n)\Gamma(n)}\) from the reflection formula.

   Rewriting this, we get:
   \[
   \Gamma(n-1) = \frac{\pi}{\sin(\pi (n-1))\Gamma(2-n)}
   \]

6. **Final Equation:**

   Using the symmetry and periodicity of the sine function, \(\sin(\pi (1-z)) = \sin(\pi z)\), the equation simplifies to:
   \[
   \Gamma(n-1) = \frac{\pi}{\sin(\pi n)\Gamma(n)}
   \]

   Therefore, we have:
   \[
   \Gamma(n)\Gamma(n-1) = \frac{\pi}{\sin(\pi n)}
   \]

Thus, the derivation of \(\Gamma(n)\Gamma(n-1) = \frac{\pi}{\sin(n\pi)}\) is complete.

Do you want details or have any questions?

Here are some related questions:
1. What is the Gamma function and how is it defined?
2. What are the properties of the Gamma function?
3. How does Euler's reflection formula relate to the Gamma function?
4. Can you derive the recurrence relation for the Gamma function?
5. What is the significance of the sine function in Euler's reflection formula?
6. How does the periodicity of the sine function affect the Gamma function?
7. Can you apply the derived formula to specific values of \(n\)?
8. How does the Gamma function generalize the factorial function?

**Tip:** Understanding the properties and relations of special functions like the Gamma function can greatly simplify complex integral and product expressions in advanced mathematics.

Learn how to derive the equation Γ(n)Γ(n-1) = π/sin ηπ using properties of the Gamma function and Euler's reflection formula. This problem involves advanced mathematical concepts, including the definition of the Gamma function, Euler's reflection formula, and recurrence relations. Suitable for advanced mathematics enthusiasts.

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