Since the category is "Calculus", no doubt there must be some way
calculus can help prove the cos(sin(x)) is always greater than
Let f(x) = cos(sin(x)) - sin(cos(x))
We suspect f(x) is always positive. Let's find the minimum value of f(x).
Start by finding the first derivative, f'(x), and set it equal to zero.
f'(x) = -sin(sin(x)) * cos(x) + cos(cos(x)) * sin(x)
which I haven't been able to break down further.
But from graphing it out, f'(x) is equal to zero at a series of well defined values for x:
0, pi, 2pi, ..., k*pi and
L, 2pi + or - L, 4pi + or - L, .....
where L is approx: .6927.... something
(and probably at negative values of x too, but I didn't check that)
So, we now have a limited number of x values to check to proove that cos(sin) is always greater than sin(cos).
Not totally rigorous, partly uses numeric methods. It's enough to convince me.
Hopefully someone can provide a complete pure mathematical proof.
Posted by Larry
on 2005-08-12 22:43:02