Question

Number of positive integral values of n for which sin(5π/2n) = sin(3π/n)

Answer 1

0

Answer 2

We are asked to find the number of positive integral values of $n$ for which the equation $\sin(5\pi/2n) = \sin(3\pi/n)$ holds true.

The general solution for $\sin x = \sin y$ is given by $x = m\pi + (-1)^m y$, where $m$ is an integer. Let $x = 5\pi/2n$ and $y = 3\pi/n$. So, $5\pi/2n = m\pi + (-1)^m (3\pi/n)$ for some integer $m$. Since $n$ is a positive integer, $n \ge 1$. We can divide by $\pi$: $5/2n = m + (-1)^m (3/n)$

We consider two cases based on the parity of $m$.

Case 1: $m$ is an even integer. Let $m = 2k$ for some integer $k$. The equation becomes $5/2n = 2k + (-1)^{2k} (3/n)$. Since $(-1)^{2k} = 1$ for any integer $k$, we have: $5/2n = 2k + 3/n$ To clear the denominators, multiply the entire equation by $2n$: $5 = (2k)(2n) + (3/n)(2n)$ $5 = 4kn + 6$ $4kn = 5 - 6$ $4kn = -1$ Since $n$ is a positive integer, $n \ge 1$. If $k$ is a positive integer ($k \ge 1$), then $4kn \ge 4(1)(1) = 4$, which cannot be equal to -1. If $k = 0$, then $4(0)n = 0$, which cannot be equal to -1. If $k$ is a negative integer ($k \le -1$), let $k = -j$ where $j$ is a positive integer ($j \ge 1$). Then $4(-j)n = -1$, which simplifies to $-4jn = -1$, or $4jn = 1$. Since $j$ and $n$ are positive integers, $4jn$ must be a positive integer. For $4jn = 1$, the only possible positive integer value for $4jn$ is 1. So, $jn = 1/4$. Since $j$ and $n$ are positive integers, their product $jn$ must also be a positive integer. However, $1/4$ is not an integer. Thus, there are no integer values of $k$ for which $n$ is a positive integer in this case.

Case 2: $m$ is an odd integer. Let $m = 2k+1$ for some integer $k$. The equation becomes $5/2n = (2k+1) + (-1)^{2k+1} (3/n)$. Since $(-1)^{2k+1} = -1$ for any integer $k$, we have: $5/2n = (2k+1) - 3/n$ To clear the denominators, multiply the entire equation by $2n$: $5 = (2k+1)(2n) - (3/n)(2n)$ $5 = (2k+1)2n - 6$ Add 6 to both sides: $5 + 6 = (2k+1)2n$ $11 = (2k+1)2n$ We are looking for positive integral values of $n$. Since $n$ is a positive integer, $2n$ is a positive even integer ($2n \in \{2, 4, 6, \dots\}$). The term $(2k+1)$ is always an odd integer for any integer $k$. The product of an odd integer $(2k+1)$ and an even integer $(2n)$ is always an even integer. So, the right-hand side $(2k+1)2n$ must be an even integer. However, the left-hand side of the equation is 11, which is an odd integer. An odd integer cannot be equal to an even integer. Thus, the equation $11 = (2k+1)2n$ has no integer solutions for $k$ that would yield a positive integer solution for $n$.

From both cases, we find that there are no positive integral values of $n$ that satisfy the given equation. Therefore, the number of positive integral values of $n$ is 0.