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cos(sinx) > sin(cosx)
By Niranjan Srinivas on Friday, November 09, 2001 - 02:32 pm:

Prove that cos(sinx) > sin(cosx)
I'm stumped !!

By David Loeffler on Friday, November 09, 2001 - 03:40 pm:

Hmm. That's a really diabolically nice inequality!

I have found a somewhat fiddly proof. Can anyone find a better one?

Since both functions are periodic, we need only prove this for -Pi < x < Pi. Furthermore, if f(x) = cos(sin(x)) and g(x) = sin(cos(x)) then f(-x) = f(x) and g(-x) = g(x), so we can look only at the interval (0, Pi).

Firstly, note that since -1 < sin(x) < 1, and 1 < Pi/2, cos(sin(x)) is always positive.
However, if Pi/2 < x < Pi, sin(cos(x)) < 0.

So the only problem is the interval (0, Pi/2).

On this interval, the sine function is increasing. Since we are trying to prove that sin(cos(x)) < sin(Pi/2 - sin(x)), it suffices to prove that cos x < Pi/2 - sin(x).

However, this is true as sin(x) + cos(x) < sqrt(2) < Pi/2.

David