This question comes from the first problem set of the MIT OCW calc course. It asks for the formula of the perimeter of a regular polygon with n sides with a radius of 1. The answer sheet has this for the answer
$$n\sin(2\pi/n)$$
If each wedge of the polygon creates a chord length of $2\sin(\theta/2)$ (the half-angle formula) and each wedge is $2\pi/n$ radians why is the answer not $2n\sin(\pi/n)$? I get the $(\pi/n)$ by multiplying $2\pi/n$ by $1/2$.
I know I must be missing something, but I'm not sure where I'm making the mistake.
$\endgroup$ 21 Answer
$\begingroup$Assuming regular polygon,
$$ AB=u = R \cdot \sin \dfrac{\pi}{n} $$
perimeter
$$ 2u n = 2nR \sin \dfrac{\pi}{n} $$
when $n\to \infty $ perimeter $\to 2 \pi R$
Take $R=1$ here if so desired in this example.
It may be a typo error in the student's answer sheet.
Here the polygon drawn is for $n=8$ sides.
