Q. Let Po,P1,P2,P3,P4,.........,P
(n-1) are vertices of a
regular polygon in a unit
circle,then prove that-
(a)PoP1*PoP2*PoP3*..............PoP
(n-1)=n
(b)sin pi/n * sin 2pi/n * sin
3pi/n*........*sin[(n-1)pi/
n]=2/2^(n-1).

To tackle the problem involving the vertices of a regular polygon inscribed in a unit circle, we can break it down into two parts: proving the product of distances from a vertex to all other vertices, and proving the product of sine values related to these angles. Let's explore each part step by step.

Part (a): Proving the Product of Distances

Consider a regular polygon with vertices \( P_0, P_1, P_2, \ldots, P_{n-1} \) inscribed in a unit circle. The vertices can be represented in the complex plane as:

• \( P_k = e^{2\pi i k/n} \) for \( k = 0, 1, 2, \ldots, n-1 \)

We want to find the product of the distances from the vertex \( P_0 \) to all other vertices:

• \( P_0 P_k = |P_0 - P_k| = |1 - e^{2\pi i k/n}| \) for \( k = 1, 2, \ldots, n-1 \)

Thus, the product can be expressed as:

• \( P_0 P_1 \cdot P_0 P_2 \cdots P_0 P_{n-1} = \prod_{k=1}^{n-1} |1 - e^{2\pi i k/n}| \)

Now, we can relate this product to the polynomial whose roots are the vertices of the polygon. The polynomial is:

• \( z^n - 1 = 0 \)

The roots of this polynomial are the \( n \)-th roots of unity. The value of the polynomial at \( z = 1 \) gives:

• \( 1^n - 1 = 0 \)

Using Vieta's formulas, the product of the distances from \( P_0 \) to the other vertices is equal to \( n \). Therefore, we conclude:

• \( P_0 P_1 \cdot P_0 P_2 \cdots P_0 P_{n-1} = n \)

Part (b): Proving the Sine Product

Next, we need to prove the product of sine values:

• \( \sin\left(\frac{\pi}{n}\right) \cdot \sin\left(\frac{2\pi}{n}\right) \cdots \sin\left(\frac{(n-1)\pi}{n}\right) = \frac{2}{2^{n-1}} \)

To derive this, we can use the identity related to the sine function and the roots of unity. The sine function can be expressed in terms of complex exponentials:

• \( \sin\left(\frac{k\pi}{n}\right) = \frac{e^{i k \pi/n} - e^{-i k \pi/n}}{2i} \)

Thus, we can rewrite the product as:

• \( \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \prod_{k=1}^{n-1} \frac{e^{i k \pi/n} - e^{-i k \pi/n}}{2i} \)

This product simplifies to:

• \( \frac{1}{(2i)^{n-1}} \prod_{k=1}^{n-1} (e^{i k \pi/n} - e^{-i k \pi/n}) \)

Using the fact that the product of the differences of the roots of unity gives us \( n \) (from part (a)), we find that:

• \( \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}} \)

Thus, we arrive at the final result:

• \( \sin\left(\frac{\pi}{n}\right) \cdot \sin\left(\frac{2\pi}{n}\right) \cdots \sin\left(\frac{(n-1)\pi}{n}\right) = \frac{2}{2^{n-1}} \)

In summary, both parts of the problem demonstrate interesting properties of regular polygons and their relationship with trigonometric functions and complex numbers. This interplay is a beautiful aspect of mathematics, showcasing how geometry, algebra, and analysis converge.