Question

HELP WITH HINTS

Let S_n be a sequence of the number of sides of regular polygon with sides3*2^n inscribed in the disc with radius 1. Let a_n,b_n denote the area of regular polygon with sides S_n inscribed in the dice and circumscribed the disc, respectively.
Show that a_n=1/2 S_n sin⁡(2π/n)
b_n =S_n tan(π/S_n )

Answer 1

Tried anything ?

Answer 2

Yea
Length though
Am i allowed to attach a pics on this website?

Answer 3

Sure you can. Click "Attach File" blue button below this textbox

Answer 4

@ganeshie8 This is what i have attempted....for the prove of a_n the final answer i get comes with a denominator s_n so i think that is not right... Then i attached for b_n as well for any correction if any. Thanks

Answer 5

` Let S_n be a sequence of the number of sides of regular polygon with sides 3*2^n inscribed in the disc with radius 1. `

I don't understand the question very well. What is the number of sides ?

Answer 6

So \(S_n\) is the length of the side, when there are \(3\times 2^n\) sides ?

Answer 7

Yes

Answer 8

From my point of view when \[ n = 1 \], \[S_n becomes S_1\]

Answer 9

Sure.
But even though english is not my first language, I'm having a hard time understanding the question. As it is given, it makes no sense to me. And your answer to my question contradicts the statement..

Answer 10

Let's try. Choose \(n\).. for example \(n=1\). Then,
- there are \(3\cdot 2^1\) sides
- the length of each side is \(S_n\).

Prove that \(a_n = \frac12 S_n \sin(\frac{2\pi}n)\)?

Answer 11

angle \(\alpha_n = \frac{2n}{3\cdot 2^n}\).