19: Denoting
h(x) = sin(x) − 2x/π
one gets
h′(x) = cos(x) − 2/π.
This is equal to zero only once within the interval
M = (0,π/2),
at some c, so we know that h′ is positive on
(0,c)
and negative on
(c,π/2).
Consequently h is increasing on
[0,c]
and decreasing on
[c,π/2].
Putting this together with the fact that
h(0+) = 0
and
h((π/2)-) = 0
we obtain that
h > 0
on
(0,π/2)
as required.