Consider a regular polygon with n-sides ith radius 1. I want to construct an equation in polar coordinates for this polygon.

On the internet, I found this:

(first part) Consider the equation of just one half of one side of the $n$-gon. It's the side of a right triangle opposite angle $A = π/n$. The hypotenuse of this triangle is $1$ (the radius of the $n$-gon), so the equation of this one side is $\cos(π/n)/\cos(θ)$, where $θ$ goes from $0$ to $π/n$. The other half of this side has the same equation, so we're already at the point where we know the equation of one side of the $n$-gon: $r = \cos(π/n) / \cos(θ), -π/n ≤ θ ≤ π/n .$

(second part  )Next, we will need to develop a function of θ that "normalizes" θ to be within plus or minus $π/n$. This is accomplished by the "floor" function, as follows: $B = θ - 2 π/n \lfloor(n θ + π)/(2 π)\rfloor $

Putting it together, the equation of an n-gon is $r = \cos(A)/\cos(B)$, which is $r = \cos(π/n)\cos(θ - 2 π/n \lfloor(n θ + π)/(2 π)\rfloor) $

someone can give me a help to understand the second part or show me an another solution?

thanks in advance

Consider a regular polygon with n-sides ith radius 1. I want to construct an equation in polar coordinates for this polygon.

On the internet, I found this:

(first part) Consider the equation of just one half of one side of the $n$-gon. It's the side of a right triangle opposite angle $A = π/n$. The hypotenuse of this triangle is $1$ (the radius of the $n$-gon), so the equation of this one side is $\cos(π/n)/\cos(θ)$, where $θ$ goes from $0$ to $π/n$. The other half of this side has the same equation, so we're already at the point where we know the equation of one side of the $n$-gon: $r = \cos(π/n) / \cos(θ), -π/n ≤ θ ≤ π/n .$

(second part)Next, we will need to develop a function of θ that "normalizes" θ to be within plus or minus $π/n$. This is accomplished by the "floor" function, as follows: $B = θ - 2 π/n \lfloor(n θ + π)/(2 π)\rfloor $

Putting it together, the equation of an n-gon is $r = \cos(A)/\cos(B)$, which is $r = \cos(π/n)/\cos(θ - 2 π/n \lfloor(n θ + π)/(2 π)\rfloor) $

Someone can give me a help to understand the second part or show me an another solution?

Thanks in advance.

Consider a regular poligon with n-sides ith radius 1. I want to constructo na equation in polar coordinates for this poligon. on the internet i found this:

(first part)consider the equation of just one half of one side of the n-gon. It's the side of a right triangle opposite angle $A = π/n$. The hypotenuse of this triangle is 1 (the radius of the n-gon), so the equation of this one side is $cos(π/n)/cos(θ)$, where $θ$ goes from $0$ to $π/n$. The other half of this side has the same equation, so we're already at the point where we know the equation of one side of the n-gon: $r = cos(π/n) / cos(θ), -π/n ≤ θ ≤ π/n .$

(second part )Next, we will need to develop a function of θ that "normalizes" θ to be within plus or minus π/n. This is accomplished by the "floor" function, as follows: $B = θ - 2 π/n floor((n θ + π)/(2 π)) $

Putting it together, the equation of an n-gon is $r = cos(A)/cos(B)$, which is $r = cos(π/n)/cos(θ - 2 π/n floor((n θ + π)/(2 π))) $

someone can give me a help to understand the second part or show me an another solution?

thanks in advance

Consider a regular polygon with n-sides ith radius 1. I want to construct an equation in polar coordinates for this polygon.

On the internet, I found this:

(first part) Consider the equation of just one half of one side of the $n$-gon. It's the side of a right triangle opposite angle $A = π/n$. The hypotenuse of this triangle is $1$ (the radius of the $n$-gon), so the equation of this one side is $\cos(π/n)/\cos(θ)$, where $θ$ goes from $0$ to $π/n$. The other half of this side has the same equation, so we're already at the point where we know the equation of one side of the $n$-gon: $r = \cos(π/n) / \cos(θ), -π/n ≤ θ ≤ π/n .$

(second part )Next, we will need to develop a function of θ that "normalizes" θ to be within plus or minus $π/n$. This is accomplished by the "floor" function, as follows: $B = θ - 2 π/n \lfloor(n θ + π)/(2 π)\rfloor $

Putting it together, the equation of an n-gon is $r = \cos(A)/\cos(B)$, which is $r = \cos(π/n)\cos(θ - 2 π/n \lfloor(n θ + π)/(2 π)\rfloor) $

someone can give me a help to understand the second part or show me an another solution?

thanks in advance