Not So Simple Polygon Embedding
The statement of this problem is the same as the statement of problem C1. The only difference is that, in problem C1, $$$n$$$ is always even, and in C2, $$$n$$$ is always odd.You are given a regular polygon with $$$2 \cdot n$$$ vertices (it's convex and has equal sides and equal angles) and all its sides have length $$$1$$$. Let's name it as $$$2n$$$-gon.Your task is to find the square of the minimum size such that you can embed $$$2n$$$-gon in the square. Embedding $$$2n$$$-gon in the square means that you need to place $$$2n$$$-gon in the square in such way that each point which lies inside or on a border of $$$2n$$$-gon should also lie inside or on a border of the square.You can rotate $$$2n$$$-gon and/or the square.
The first line contains a single integer $$$T$$$ ($$$1 \le T \le 200$$$) — the number of test cases.Next $$$T$$$ lines contain descriptions of test cases — one per line. Each line contains single odd integer $$$n$$$ ($$$3 \le n \le 199$$$). Don't forget you need to embed $$$2n$$$-gon, not an $$$n$$$-gon.
Print $$$T$$$ real numbers — one per test case. For each test case, print the minimum length of a side of the square $$$2n$$$-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed $$$10^{-6}$$$.
import math
casos = int(input())
for t in range(casos):
    n = int(input())
    angulo = math.pi / (4 * n)
    print(1 / math.sin(angulo) / 2)
import math
t = int(input())
while t > 0:
    t -= 1
    n = int(input())
    b = math.pi * (1 / 4 + (n - 1) // 2 / 2 / n)
    print(math.sin(b) / math.sin(math.pi / 2 / n))
from math import sin, cos, tan, pi
for _ in range(int(input())):
    n = int(input())
    if n % 2 == 0:
        print(1 / tan(pi / (2 * n)))
    else:
        print(cos(pi / (4 * n)) / sin(pi / (2 * n)))