www.batmath.it

Consider a right angled triangle OAB, with hypotenuse OB=1, and
BOA=*x*. Then the area of the circular segment OAD is
less than the area of the triangle OAB which is less than that
of the circular segment OCB. As OA=cos*x*,
BA=sin*x*, we obtain:

.

If we divide by (1/2)*x*cos*x*, we obtain:

.

If we change *x* with *-x*, nothing changes. So we
may conclude.

The three graphs are plotted here below.

copyright 2000 et seq. maddalena falanga & luciano battaia

first published on march 26 2002 - last updated on september 01
2003