$\pi$, Dedekind cuts, trigonometric functions, area of a circle

Solution 1:

The following answers your first question. We do not mention sine or cosine. We will use some basic integration techniques, but of course we will not use trigonometric substitution!

Define the circle of radius $r$ with centre the origin by the equation $x^2+y^2=r^2$, and the disk by $x^2+y^2\le r^2$.

We first show that the area of a disk of radius $r$ is a constant times $r^2$. By symmetry the area is $$4\int_0^r\sqrt{r^2-x^2}\,dx.$$ Make the change of variable $x=rt$. Our area is $$r^2\left(4\int_0^1 \sqrt{1-t^2}\,dt\right).$$ Thus $4\int_0^1 \sqrt{1-t^2}\,dt$ is the desired constant. We could call it $\pi$. But let us call it $4k$.

For the circumference, the usual arclength formula, after some simplification, gives that the circumference is $$4\int_0^r \frac{r\,dx}{\sqrt{r^2-x^2}}.$$ The change of variable $x=rt$ transforms this to $$r\left(4\int_0^1 \frac{dt}{\sqrt{1-t^2}}\right).$$ So we have $r$ times a constant. Which constant?

We will evaluate $\int_0^1\sqrt{1-t^2}\,dt$ (yes!) in a funny way, by parts. Let $u=\sqrt{1-t^2}$ and $dv=dt$. Then $du=-\frac{t}{\sqrt{1-t^2}}\,dt$ and $v=t$. After a little while we find that $$\int_0^1\sqrt{1-t^2}\,dt=\int_0^1 \frac{t^2\,dt}{\sqrt{1-t^2}}.$$ But the numerator on the right is $1-(1-t^2)$. And $\dfrac{1-t^2}{\sqrt{1-t^2}}=\sqrt{1-t^2}$. Thus $$\int_0^1\sqrt{1-t^2}\,dt=\int_0^1 \frac{dt}{\sqrt{1-t^2}}-\int_0^1\sqrt{1-t^2}\,dt.$$ We conclude that $$\int_0^1 \frac{dt}{\sqrt{1-t^2}}=2\int_0^1\sqrt{1-t^2}\,dt=2k.$$ It follows that the circumference of a circle of radius $r$ is $8kr$.

Remark: One can introduce the trigonometric functions via integrals, as mentioned in the OP. Then their basic properties, such as the addition laws, are not difficult to derive. It is, however, mildly tedious.

Solution 2:

If you believe in limits, Archimedes' approach to the calculation of $\pi$ satisfies the first bullet. This is one definition of $\pi$.

For the second bullet, you just need the ability to compare any two rationals. Then follow through Archimedes' calculation and eventually any rational will be one side of $\pi$ or the other. Completeness says that $\pi$ exists.