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Next: 9.3 Integrals of the Up: 9. Trigonometric Functions Previous: 9.1 Properties of Sine   Index

9.2 Calculation of $\pi$

The proof of the next lemma depends on the following assumption, which is explicitly stated by Archimedes [2, page 3]. This assumption involves the ideas of curve with given endpoints and length of curve (which I will leave undiscussed).

9.32   Assumption. Let $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ be points in $\mbox{{\bf R}}^2$. Then of all curves with endpoints $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$, the segment $[\mbox{{\bf a}}\mbox{{\bf b}}]$ is the shortest.

9.33   Lemma.

\begin{displaymath}\sin (x)<x \mbox{ for all } x \in \mbox{${\mbox{{\bf R}}}^{+}$},\end{displaymath}


\begin{displaymath}\vert\sin (x)\vert\leq \vert x\vert \mbox{ for all } x\in\mbox{{\bf R}}.\end{displaymath}


Case 1: Suppose $\displaystyle {0<x<{\pi\over 2}}$.

Then (see the figure) the length of the arc joining $W(-x)$ to $W(x)$ in the first and fourth quadrants is $x+x=2x$. (This follows from the definition of $W$.) The length of the segment $[W(x)W(-x)]$ is $2\sin
(x)$. By our assumption, $2\sin(x)\leq 2x$, i.e., $\sin(x)\leq x$. Since both $x$ and $\sin (x)$ are positive when ${0<x<{\pi\over 2}}$, we also have $\vert\sin (x)\vert\leq \vert x\vert$.

Case 2: Suppose $\displaystyle {x\geq {\pi\over 2}}$. Then

\begin{displaymath}\sin (x)\leq \vert\sin (x)\vert\leq 1<{\pi\over 2}\leq x=\vert x\vert\end{displaymath}

so $\sin(x)\leq x$ and $\vert\sin (x)\vert\leq \vert x\vert$ in this case also. This proves the first assertion of lemma 9.33. If $x<0$, then $-x>0$, so

\begin{displaymath}\vert\sin (x)\vert=\vert\sin (-x)\vert\leq \vert-x\vert=\vert x\vert.\end{displaymath}


\begin{displaymath}\vert\sin (x)\vert\leq \vert x\vert \mbox{ for all } x\in\mbox{{\bf R}}\setminus \{0\},\end{displaymath}

and since the relation clearly holds when $x=0$ the lemma is proved. $\diamondsuit$

9.34   Lemma (Limits of sine and cosine.) Let $a\in\mbox{{\bf R}}$. Let $\{a_n\}$ be a sequence in $\mbox{{\bf R}}$ such that $\{a_n\}\to a$. Then

\begin{displaymath}\{\cos (a_n)\}\to\cos (a) \mbox{ and }\{\sin (a_n)\}\to\sin

Proof: By (9.25) we have

\begin{displaymath}\cos (a_n)-\cos (a)=-2\sin\Big( {{a_n+a}\over 2}\Big)\sin\Big( {{a_n-a}\over


0 \leq \vert\cos (a_n)-\cos (a)\vert &\leq& 2\vert
\sin \Big( ...
... 2}\Big)\vert\leq 2\vert {{a_n-a}\over 2}\vert=\vert a_n-a\vert.

If $\{a_n\}\to a$, then $\{ \vert a_n-a\vert\} \to 0$, so by the squeezing rule,

\begin{displaymath}\{\vert\cos(a_n) - \cos(a)\vert\}\to 0.\end{displaymath}

This means that $\{ \cos (a_n) \}
\to \cos(a)$.

The proof that $\{\sin (a_n)\}\to\sin (a)$ is similar. $\diamondsuit$

The proof of the next lemma involves another new assumption.

9.35   Assumption. Suppose $\displaystyle {0<x<{\pi\over 2}}$. Let the tangent to the unit circle at $W(x)$ intersect the $x$ axis at $\mbox{{\bf p}}$, and let $\mbox{{\bf q}}=(1,0)$.
Then the circular arc joining $W(x)$ to $W(-x)$ (and passing through $\mbox{{\bf q}}$) is shorter than the curve made of the two segments $[W(x)\mbox{{\bf p}}]$ and $[\mbox{{\bf p}}W(-x)]$ (see the figure).

Remark: Archimedes makes a general assumption about curves that are concave in the same direction [2, pages 2-4] which allows him to prove our assumption.

9.36   Lemma. If $\displaystyle {0<x<{\pi\over 2}}$, then

\begin{displaymath}x \leq {\sin(x)\over \cos(x)}. \end{displaymath}

Proof: Suppose $\displaystyle {0<x<{\pi\over 2}}$. Draw the tangents to the unit circle at $W(x)$ and $W(-x)$ and let the point at which they intersect the $x$-axis be $\mbox{{\bf p}}$. (By symmetry both tangents intersect the $x$-axis at the same point.) Let $\mbox{{\bf b}}$ be the point where the segment $[W(x)W(-x)]$ intersects the $x$-axis, and let $\mbox{{\bf r}}=W(x)$. Triangles $\mbox{{\bf0}}
\mbox{{\bf b}}\mbox{{\bf r}}$ and $\mbox{{\bf0}} \mbox{{\bf r}}\mbox{{\bf p}}$ are similar since they are right triangles with a common acute angle.


\begin{displaymath}{{\mbox{distance}(\mbox{{\bf r}},\mbox{{\bf b}})}\over {\mbox...
...\bf r}})}\over {\mbox{distance}(\mbox{{\bf0}},\mbox{{\bf r}})}}\end{displaymath}


\begin{displaymath}{{\sin (x)}\over {\cos (x)}}= {{\mbox{distance}(\mbox{{\bf p}},\mbox{{\bf r}})}\over 1}.\end{displaymath}

Now the length of the arc joining $W(x)$ to $W(-x)$ is $2x$, and the length of the broken line from $\mbox{{\bf r}}$ to $\mbox{{\bf p}}$ to $W(-x)$ is $\displaystyle {2\Big(\mbox{distance}(\mbox{{\bf p}},\mbox{{\bf r}})\Big)=2{{\sin (x)}\over {\cos(x)}}}$, so by assumption 9.35,

\begin{displaymath}2x\leq 2{{\sin(x)}\over {\cos x}}\end{displaymath}


\begin{displaymath}x \leq {\sin(x)\over \cos(x)}. \end{displaymath}

This proves our lemma. $\mbox{ $\diamondsuit$}$

9.37   Theorem. Let $\{x_n\}$ be any sequence such that $x_n\neq 0$ for all $n$, and $\{x_n\} \to 0$. Then
\Big\{ {{\sin (x_n)}\over {x_n}}\Big\}\to 1.
\end{displaymath} (9.38)

Hence if $\sin(x_n)\neq 0$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ we also have

\begin{displaymath}\Big\{ {{x_n}\over {\sin(x_n)}}\Big\}\to 1.\end{displaymath}

Proof: If $x \in (0, {\pi\over 2})$, then it follows from lemma(9.36) that $\displaystyle {\cos(x) \leq {\sin(x) \over x} }$. Since

\begin{displaymath}\cos(-x) = \cos(x) \mbox{ and }{\sin(-x) \over -x} = {\sin(x) \over x},\end{displaymath}

it follows that

\begin{displaymath}\cos(x) \leq {\sin(x) \over x} \mbox{ whenever } 0 < \vert x\vert < {\pi \over 2}.\end{displaymath}

Hence by lemma 9.33 we have
\begin{displaymath}\cos(x) \leq {\sin(x) \over x} \leq 1
\mbox{ whenever } 0 < \vert x\vert < {\pi \over 2}.
\end{displaymath} (9.39)

Let $\{x_n\}$ be a sequence for which $x_n\neq 0$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ and $\{x_n\} \to 0$. Then we can find a number $N\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that for all $\displaystyle {n\in\mbox{{\bf Z}}_{\geq N}
(\vert x_n\vert<{\pi\over 2}})$. By (9.39)

\begin{displaymath}n\in\mbox{{\bf Z}}_{\geq N}\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\cos(x_n)\leq {{\sin(x_n)}\over {x_n}}\leq 1.\end{displaymath}

By lemma 9.34, we know that $\{\cos (x_n)\}\to 1$, so by the squeezing rule $\displaystyle { \Big\{ {{\sin(x_n)}\over {x_n}} \Big\}\to 1}$. $\diamondsuit$

9.40   Example (Calculation of $\pi$.) Since $\displaystyle { \Big\{ {\pi\over n}\Big\} \to 0 }$, it follows from (9.38) that

\begin{displaymath}\lim\Big\{ {{\sin \Big( {\pi\over n}\Big)}\over { {\pi\over n} }}\Big\}=1\end{displaymath}

and hence that

\begin{displaymath}\lim\Big\{n\sin\Big( {\pi\over n}\Big)\Big\}=\pi.\end{displaymath}

This result can be used to find a good approximation to $\pi$. By the half-angle formula, we have

\begin{displaymath}\sin^2 \Big( {t\over 2}\Big) ={{1-\cos t}\over 2}={1\over

for $0\leq t\leq \displaystyle { {\pi\over 2}}$. Here I have used the fact that $\cos t\geq
0$ for $0\leq t\leq \displaystyle { {\pi\over 2}}$. Also $\displaystyle { \sin ({\pi\over 2})=1}$ so

\sin^2({\pi\over 4})&=&{1\over 2}\Big(1-\sqrt{1-\sin^2{\pi\ove...
...sqrt{1-{1\over 2}}\Big)={1\over 2}\Big(1-\sqrt{{1\over 2}}\Big).

By repeated applications of this process I can find $\displaystyle {\sin^2\Big( {\pi\over
{2^n}}\Big)}$ for arbitrary $n$, and then find

\begin{displaymath}2^n\sin \Big( {\pi\over {2^n}}\Big)\end{displaymath}

which will be a good approximation to $\pi$.

I wrote a set of Maple routines to do the calculations above. The procedure sinsq(n) calculates $\displaystyle {\sin^2\Big( {\pi\over
{2^n}}\Big)}$ and the procedure mypi(m) calculates $\displaystyle { 2^m\sin\Big( {\pi\over {2^m}}\Big)}$. The `` fi" (which is `` if" spelled backwards) is Maple's way of ending an `` if" statement. `` Digits $:= 20$" indicates that all calculations are done to $20$ decimal digits accuracy. The command `` evalf(Pi)" requests the decimal approximation to $\pi$ to be printed.

> sinsq := 
>    n-> if n=1 then 1;
>        else .5*(1-sqrt(1 - sinsq(n-1)));
>        fi;
sinsq := proc(n) options operator,arrow; if n = 1 then 1 else .5 -.5*sqrt(1-sinsq(n-1)) fi end

> mypi := m -> 2^m*sqrt(sinsq(m));

{\it mypi} := {m} \rightarrow 2^{{m}} {\rm sqrt}( {\rm sinsq}(
 {m} ) )

> Digits := 20;

{\it Digits} := 20

> mypi(4);


> mypi(8);


> mypi(12);


> mypi(16);


> mypi(20);


> mypi(24);


> mypi(28);


> mypi(32);


> mypi(36);


> mypi(40);


> evalf(Pi);


9.41   Exercise. Examine the output of the program above. It appears that $\pi=0$. This certainly is not right. What can I conclude about $\pi$ from my computer program?

9.42   Exercise. Show that the number $\displaystyle { n\sin\Big( {\pi\over n}\Big)}$ is the area of a regular $2n$-gon inscribed in the unit circle. Make any reasonable geometric assumptions, but explain your ideas clearly.

next up previous index
Next: 9.3 Integrals of the Up: 9. Trigonometric Functions Previous: 9.1 Properties of Sine   Index
Ray Mayer 2007-09-07