My definition for $\pi$ is twice the first positive real number such that $\cos(x)=0$.

I think it's not even feasible to evaluate that $3.14 <\pi < 3.15$ in elementary level.

Well, the only tool i've learned is the Gamma function. (Such as $\int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$,$\Gamma(1/2)=\sqrt{\pi}$)

I don't want really much but at least i want to show that $3<\pi<4$.

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    $\begingroup$ $\pi < 4$ because the area of an unit circle is smaller than that of a $2 \times 2$ square. $\endgroup$ Commented Feb 24, 2014 at 7:26
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    $\begingroup$ How does the accepted answer fit in your definition for $\pi$? If every approach was allowed, why did you state what your definition for $\pi$ was? $\endgroup$
    – Did
    Commented Feb 24, 2014 at 7:54
  • $\begingroup$ @Did: I added an alternative answer concomitant with the OPs definition of $\pi$. $\endgroup$
    – copper.hat
    Commented Feb 25, 2014 at 3:45
  • $\begingroup$ Here is a proof using two integrals similar to the one for $\frac{22}{7}-\pi>0$ math.stackexchange.com/a/1618454/134791 (Is this question a duplicate?) $\endgroup$ Commented Mar 3, 2016 at 22:23

2 Answers 2


When $x>0$, the Taylor series for $\cos x$ is alternating and eventually decreasing in absolute value, so we can get an upper bound for it by truncating after any positive term after it starts decreasing, and a lower bound by truncating after any negative term after it starts decreasing.

In particular, $\cos 2 < 1 - \frac{2^2}{2!} + \frac{2^4}{4!} = -\frac{1}{3}$.

Similarly, for $0 \leq x \leq 3/2$, $\cos x > p(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}$. I claim that $p$ is positive on this interval. To see this, note that $p'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!} < -x + \frac{x^3}{6}$, which is negative on the interval $[0, \sqrt{6}]$, so $p$ is decreasing on $[0, \sqrt{6}]$. Also, $p(3/2) = \frac{359}{5120} > 0$; thus $p$ is positive on all of $[0, 3/2]$.

It follows by the intermediate value theorem that $\cos x$ has a zero between $3/2$ and $2$. Since $\cos x$ is positive on $[0, 3/2]$, this is its first positive zero; thus $3 < \pi < 4$.


Inscribe a hexagon of side 1 inside a circle of radius 1. Then the triangle inequality (with limits) shows that $6 < 2 \pi$.

There is an elementary proof (see Wikipedia) that $\pi < { 22 \over 7}$ (which is $<4$).

Alternative answer: (I feel the need to redeem myself :-).)

Micah's answer is far more succinct, but I like the following approach as it forges a link between the analytical and geometric aspects.

I am assuming that $\cos $ (and $\sin$) are defined using the usual power series, and $\pi$ is defined as in the question. Since $\sin' = \cos$, we see that $\sin$ is increasing (and hence positive) on $(0,{ \pi \over 2}]$.

By differentiating the function $x \mapsto \cos^2 x+ \sin^2 x$ and evaluating at $x=0$ we can establish the identity $\cos^2 x+ \sin^2 x = 1$. (This gives $\sin {\pi \over 2} = 1$.)

Similarly, by differentiating the function $x \mapsto (\cos x -\sin({\pi \over 2} -x))^2 + (\sin x -\cos({\pi \over 2} -x))^2$ we see that it is constant and by evaluating at $x=0$, we obtain the identities $\cos x =\sin({\pi \over 2} -x)$ and $\sin x =\cos({\pi \over 2} -x)$.

From this we get $\cos { \pi \over 4} = \sin { \pi \over 4} = {1 \over \sqrt{2}}$.

If $\gamma:[0,1] \to \mathbb{R}^2$ is $C^1$, we define the length as $l(\gamma) = \sup_{{\cal P}} \sum_k | \gamma(t_{k+1}) -\gamma(t_k)|$, where ${\cal P}$ are the partitions of $[0,1]$ and we note that $l(\gamma) = \int_0^1 \|\gamma'(t)\| dt$.

Now consider the curves $\lambda(t) = (1,0) + t ({1 \over \sqrt{2}}-1, {1 \over \sqrt{2}})$, $\gamma(t) = (\cos (t { \pi \over 4}), \sin (t { \pi \over 4}))$, and $\upsilon(t) = (1,0)+t (0,1)$.

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We have $\|\lambda'(t)\| = \sqrt{2 -\sqrt{2}}$, $\|\gamma'(t)\| = {\pi \over 4}$, and $\|\upsilon'(t)\| =1$.

Note that for the straight lines $\lambda, \upsilon$, for any partition $(0=t_1,...,t_n=1)$ we have $l(\lambda) = \sum_k |\lambda(t_{k+1})-\lambda(t_k)|$, and similarly for $\upsilon$. In particular, for any strictly increasing continuous function $\eta:[0,1] \to [0,1]$ with $\eta(0) = 0, \eta(1) = 1$, we have $l(\upsilon) = l(\upsilon \circ \eta)$ (this is true more generally, of course).

We have $\lambda(0) = \gamma(0), \lambda(1) = \gamma(1)$, and taking any partition of the form $(0,t,1)$ with $t \in (0,1)$ we have $l(\gamma) = |\lambda(1)-\lambda(0)| = |\gamma(1)-\gamma(0)| < |\gamma(1)-\gamma(t)| + |\gamma(t)-\gamma(0)| \le l(\gamma)$, and so $\sqrt{2 -\sqrt{2}} < {\pi \over 4}$. A little manipulation shows that $4(\sqrt{2 -\sqrt{2}}) >3$, which gives $\pi > 3$.

For the other bound, let $\eta(t) = {\sin (t { \pi \over 4}) \over \cos (t { \pi \over 4})}$, then we have $\eta'(t) = {\pi \over 4} {1 \over \cos ^2(t { \pi \over 4})}>0$, and $\|(\upsilon \circ \eta)'(t) \| = {\pi \over 4} {1 \over \cos ^2(t { \pi \over 4})}$. In particular, $\|\gamma'(t)\| < \|(\upsilon \circ \eta)'(t) \|$ for all $ t \in (0,1]$. It follows that $l(\gamma) < l(\upsilon \circ \eta) = l(\upsilon)$, from which we get $\pi < 4$.


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