Ok so a cool video idea is to how this proof I thought of for proving the limit in the title. I’m only a calc 2 student so if there’s any holes in my proof let me know!

Consider a regular polygon, P, with n sides and a radius of 1. In this case, radius meaning the distance from the center of P to one of its vertices.

You can divide P into n similar triangles of equal area by connecting each vertex of P to the center.

Now, consider one of those triangles. Two of the sides are radii of P, so the triangle is isosceles. The non-similar angle is equal to 2pi/n, as in order for all the triangles to be similar, the angles must be equal and those angles must also add up to 2pi. Now, you can create two equal right triangles by splitting the original triangle exaxtly in two from the angle 2pi/n. As there are two of these right triangles per original triangles in P, there are 2n of the right trangles in P.

Now, consider one of these right triangles. The hypotenuse is 1 since that is a radius of P. One of the angles in this right triangle is pi/n, since the 2pi/n angle was split exactly in half above. In order to get the area of this triangle, you multiply together the two non-hypotenuse sides (the base, b, and the height, h) and divide by 2.

Now, we know that sin(pi/n)=b/1=b and cos(pi/n)=h/1=h. You get these by just applying the definition of sin as opposite/hypotenuse and cos as adjacent/hypotenuse respectively. Therefore, an expression for the area of one of these right triangles (A_r) is: A_r=sin(pi/n)cos(pi/n)/2

Doing some manipulation yields:

sin(pi/n)cos(pi/n)/2=2sin(pi/n)cos(pi/n)/4=sin(2pi/n)/4

So A_r=sin(2pi/n)/4

Recall that, as stated above, there are 2n of these triangles in P. Therefore, multiplying A_r by 2n will yield the area of P (A_p).

A_p=2n*A_r=2n sin(2pi/n)/4=n sin(2pi/n)/2

So A_p=n sin(2pi/n)/2

Next, we’re going to want to take the limit of A_p as n -> infinity. First, let’s consider what that actually means for our polygon, P. As the number of sides of a regular polygon approaches infinity, the polygon approaches a circle. In particula, since the radius of P is fixed at 1, as the number of sides of P approach infinity, P approaches the unit circle. Therefore, its area, by definition, approaches pi.

With this knowledge, we can now say that the limit as n->infinity of n sin(2pi/n) / 2 =pi. We could prove that limit with L’Hopital’s Rule, but we need the limit in the title to find the derivative of sin, so we really can’t use L’Hopital’s Rule in this case. Anyway, using the constant rule of limits, we can move the 1/2 to the outside of the limit. We can also multiply both sides of the above equation by 2, yielding:

limit n->infinity n sin(2pi/n) = 2pi

To pretty much finish this, we can make a substitution in the above limit. We can let x=2pi/n. Therefore, as n->infinity, x->0. Also, n=2pi/x. Substituting all of this in, we get: limit x->0 2pi/x sin(x) = 2pi. Using the constant rule of limits, we can briby the 2pi to the outside of the limit and, since 2pi is not equal to 0, we can divide it on both sides. This gives us:

limit x->0 sin(x)/x = 1

as desired.

Anyway, I hope BlackPenRedPen considers this for a video because I think this is quite cool. Hope you guys liked this!

Try this right triangle is pi/n, since the 2pi/n angle it will result split exactly in half above.

Professional Laser Skin Care Clinic In Edmonton ABGood work. So nice to see another approach.

Nice proof!

This is awesome!! BPRP should do this in his 100 trig identities video. BTW, if you make a youtube channel someday (or if you already have one) this'll get you some subs :) (at least my sub)