Regular $n$-gon sidelength

Question

I want to derive a side length for a regular $n$-gon inscribed in a unit circle. So, starting with $n=3$, I assigned the first point to $(1,0)$ and going counter clockwise, let the next 2 points be $\left(\cos\left(\frac{2\pi}{3}\right),\sin \left(\frac{2\pi}{3}\right)\right)$ and $\left(\cos\left(\frac{4\pi}{3}\right),\sin \left(\frac{4\pi}{3}\right)\right)$ respectively. After a few repetitions of this with other values of $n$, I found that the vertices of this particular $n$-gon were at $P_n=\left(\cos\left(\frac{(2n-2)\pi}{n}\right),\sin \left(\frac{(2n-2)\pi}{n}\right)\right),\quad n=1,2,\ldots,n.$

To get the sidelength, I use the distance formula and compute $\left\Vert P_2-P_1 \right\Vert$. That resulted in the following algebra: $$\left[\left( \cos\left( 2\frac{\pi}{n} \right)-1\right)^2 + \sin^2\left(2\frac{\pi}{n} \right) \right]^{1/2}$$ $$\left[ \cos^2\left( 2\frac{\pi}{n} \right) -2\cos\left( 2\frac{\pi}{n} \right)+1+ \sin^2 \left( 2\frac{\pi}{n} \right) \right]^{1/2}$$ $$\text{substitute } 1-\sin^2\left(2\frac{\pi}{n}\right)=\cos^2\left(2\frac{\pi}{n}\right)$$ $$\left[ 1-\sin^2\left( 2\frac{\pi}{n} \right) -2\cos\left( 2\frac{\pi}{n} \right)+1+ \sin^2 \left( 2\frac{\pi}{n} \right) \right]^{1/2}$$ $$\text{sidelength} = \left[ 2-2\cos\left(2\frac{\pi}{n}\right) \right]^{1/2}$$

When I look in Wolfram, I see the sidelength for $n$-gons in this unit circle given as $$\text{sidelength}=2\cdot \sin\frac{\pi}{n}$$ obviously with a different trig function and no square root. These two functions are not equivalent. Graphing them will convince you in an instant that they are identicle for $n\ge 1$ and wildly different for $n<1$. I would like some help in deriving the Wolfram sidelength result. I have not been able to algebraically/trig convert mine to his and suspect that it isn't possible since they are not even equal throughout.

Answer 1

You are only concerned with $n \gt 1$ where they agree. Note the factor $2$ in the argument of the cosine in your expression. Then using the double angle identity we have $$\begin {align} \left[ 2-2\cos\left(2\frac{\pi}{n}\right) \right]^{1/2}&=\left[ 2-2\left(1-2\sin^2\left(\frac{\pi}{n}\right) \right) \right]^{1/2}\\ &=\left[ 4\sin^2\left(\frac{\pi}{n} \right) \right]^{1/2}\\ &=2\sin \left(\frac{\pi}{n} \right) \end {align}$$ When the sine is positive, as it is here.

Answer 2

$$ 2 - 2 \cos (2x) = \left( 2\sin x \right)^2 $$ This is a standard trigonometric identity, usually written like this: $$ \frac 1 2 - \frac 1 2 \cos(2x) = \sin^2 x $$ It is often called the sine half-angle formula.