# Pi/Zero of cosine/Introduction/Section

< Pi

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The number ${\displaystyle {}\pi }$ is the area and half of the circumference of a circle with radius ${\displaystyle {}1}$. But, in order to build a precise definition for this number on this, we would have first to establish measure theory or the theory of the length of curves. Also, the trigonometric functions have an intuitive interpretation at the unit circle, but also this requires the concept of the arc length. An alternative approach is to define the functions sine and cosine by their power series, and then to define the number ${\displaystyle {}\pi }$ with the help of them, and establishing finally the relation with the circle.

## Lemma

The cosine function has, within the real interval ${\displaystyle {}[0,2]}$, exactly one zero.

### Proof

We consider the cosine series

${\displaystyle {}\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}\,.}$

For ${\displaystyle {}x=0}$, we have ${\displaystyle {}\cos 0=1}$. For ${\displaystyle {}x=2}$, one can write

{\displaystyle {}{\begin{aligned}\cos 2&=1-{\frac {2^{2}}{2!}}+{\frac {2^{4}}{4!}}-{\frac {2^{6}}{6!}}+{\frac {2^{8}}{8!}}-\ldots \\&=1-{\frac {2^{2}}{2!}}{\left(1-{\frac {4}{3\cdot 4}}\right)}-{\frac {2^{6}}{6!}}{\left(1-{\frac {4}{7\cdot 8}}\right)}-\ldots \\&=1-2(2/3)-\ldots \\&\leq -1/3.\end{aligned}}}

Hence, due to the intermediate value theorem, there exists at least one zero in the given interval.
To prove uniqueness, we consider the derivative of cosine, which is

${\displaystyle {}\cos 'x=-\sin x\,,}$

due to fact. Hence, it is enough to show that sine is positive in the interval ${\displaystyle {}]0,2[}$, because then cosine is strictly decreasing by fact in the interval and there is only one zero. Now, for ${\displaystyle {}x\in {]0,2]}}$, we have

{\displaystyle {}{\begin{aligned}\sin x&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\ldots \\&=x{\left(1-{\frac {x^{2}}{3!}}\right)}+{\frac {x^{5}}{5!}}{\left(1-{\frac {x^{2}}{6\cdot 7}}\right)}+\ldots \\&\geq x{\left(1-{\frac {4}{3!}}\right)}+{\frac {x^{5}}{5!}}{\left(1-{\frac {4}{6\cdot 7}}\right)}+\ldots \\&\geq x/3\\&>0.\end{aligned}}}
${\displaystyle \Box }$

## Definition

Let ${\displaystyle {}s}$ denote the unique (according to fact) real zero of the cosine function in the interval ${\displaystyle {}[0,2]}$. Then the number ${\displaystyle {}\pi }$ is defined by

${\displaystyle {}\pi :=2s\,.}$