definition of $\pi$ in terms of $\cos$

Question

I have to define $\pi$ in terms of cos and to show why the definition is well-defined.

I think that I haven't understood what it is asking because $\pi$ would be the angle that divided for $2$ gives me $0$ to me.

But it is clearly far from the purpose of the exercise.

Can someone help me to understand?

Answer 1

You can define $\pi$ as the smallest real number greater than or equal to $0$ such its cosine is equal to $-1$.

Answer 2

A reasonable definition: $$\pi = 2\sup\{x > 0\,\vert\,\forall t\in[0,x]:\,\cos(t) > 0\}.$$

Answer 3

I think that I haven't understood what it is asking because $\pi$ would be the angle that divided for 2 gives me 0 to me.

But it is clearly far from the purpose of the exercise.

What you suggest here is indeed (except that your phrasing is a bit sloppy) a common definition of $\pi$ and I don't think you need to assume it is not exactly the definition the exercise asks for.

As has already been pointed out, there's plenty of things you still need to do to prove that this definition actually defines something.

You could also do what José Carlos Santos suggests and look for solutions to $\cos(x)=-1$ instead of $\cos(x)=0$.

But immediately it seems to be easier to prove that $\cos(x)=0$ must have a solution in the correct range -- for example, you can evaluate $\cos(1)$ and $\cos(2)$ using the power series to sufficient precision that you can appeal to the Intermediate Value Theorem to show there must be a zero between them.

Answer 4

Let's consider $\displaystyle \cos(x)=\sum\limits_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!}$ and $\displaystyle \sin(x)=\sum\limits_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$

We can show both have infinite radius of convergence and that they are each other derivatives (with proper signs).

This allows to get also $(\cos(x)^2+\sin(x)^2)'=-2\cos(x)\sin(x)+2\sin(x)\cos(x)=0$ so the sum is constant and equal to $\cos(0)^2+\sin(0)^2=1$

Since $\cos(0)=1>0$ let's make the hypothesis that $\cos$ has no zero, i.e. $\forall x,\ \cos(x)>0$.

The derivatives relation implies that $\sin(x)\nearrow$ on $[0,+\infty)$

Since $\sin$ is increasing then:

• $\sin(1)>\sin(0)=0$
• $\displaystyle \cos(1)-\cos(x)=\int_1^x \sin(t)\mathop{dt}\ge \int_1^x \sin(1)\mathop{dt}\ge (x-1)\sin(1)\to +\infty$

We get $\cos(x)\to -\infty$ which is a contradiction with the hypothesis that $\cos(x)$ is always positive.

Thus $\cos$ annulates on $[0,+\infty)$ and we define $\dfrac{\pi}{2}$ as its first positive zero.

(it exists and is uniquely defined as minimum of a non empty, closed and bounded inferiorly real set $E=\{x\ge 0\mid \cos(x)=0\}$)

Then I let you go on verifying that $\sin(\frac\pi 2)=1$, and that $2\pi$ is the smallest period of $\sin,\cos$, both results obtainable from the derivatives, and the trigonometric addition formulas.

Answer 5

$\pi$ can be written in terms of $\cos$ as follows:

$$\pi = \lim_{n \to \infty} 2^{n+1/2}\sqrt {1-\cos\left(\left(\frac{180}{2^n}\right)^\circ\right)}$$

Answer 6

I think the best definition is \begin{equation} \pi = 2\int_0^1 \frac{d x}{\sqrt{1 - x^2}} \end{equation} or perhaps \begin{equation} \pi = \int_{-\infty}^\infty \frac{d x}{1 + x^2} \end{equation}