Integral calculus question, using limits.

[stripes]

Homework Statement

Let A be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2pi/n, show that A = (\frac{1}{2})nr^{2}sin(2(\pi)/n)

Homework Equations

Area of a circle = pi x r^2
Area of triangle = (base x height)/2
sin x = opposite/hypotenuse

The Attempt at a Solution

My attempt is in the attached word document. I spent over an hour trying to use the latex/equation stuff, and normally I know how to use it, but it kept putting in stuff that I wasn't actually typing. Every line would just show up as something random.

I apologize in advance.

 

[HallsofIvy]

Don't use the Pythagorean theorem to get h in terms of r and n. Instead, note that you have a right triangle with angle \theta/2, "near side" h and "hypotenuse" r. cos(\theta/2)= h/r so h= r cos(\theta/2). Similarly, the base of the triangle is r sin(\theta/2) so the area of each right triangle is (1/2)r^2 sin(\theta/2)cos(\theta/2).

Since sin(2x)= 2sin(x)cos(x) for any x, sin(\theta/2)cos(\theta/2)= (1/2)sin(\theta). And since each of the triangles having a side of the polygon as base is two of those right triangles, the area of each such triangle is (1/2)r^2 sin(\theta)[/tex].<br /> <br /> Now use the fact that there are n such triangles and that \theta= 2\pi/n

 

[stripes]

makes perfect sense! Then all I need to do is multiply (1/2)r^2 sin(\theta) by n and substitute \theta= 2\pi/n.

But, why wasn't I able to use Pythagorean theorem to find h?

 

[HallsofIvy]

Well, you could, but you clearly want everything in terms of \theta. Writing h in terms of other things you don't know doesn't help!