Trigonometry Inequality Involving Powers of Sin, Cos.

Question

Prove that for $0 < x < \frac{\pi}4$,

$$ \sin x^ {\sin x} < \cos x ^{\cos x}. $$

I don't have any nice ideas. I was thinking about taking the natural log and looking at the taylor series of both sides but it is too ugly.

Answer 1

How about using the monotonicity of the function $f(t)=t^{t}$ for $t>0$. Note that the function increases for $t>1/e$.

Answer 2

If you consider the function $$f(x)= \sin x^ {\sin x}-\cos x ^{\cos x}$$ its values are equal to $0$ at the bounds. If you compute the first and second derivatives (you need to use limits), you will find $$f'(0)=-\infty$$ $$f'(\frac{\pi}{4})=2^{\frac{1}{4} \left(-2-\sqrt{2}\right)} (2-\log (2))\simeq 0.723238$$ $$f''(0)=\infty$$ $$f''(\frac{\pi}{4})=0$$ Assuming that there is not other root to $f(x)=0$, then the second derivative is positive and the inequality is satisfied.

You could show that $f(x)$ goes though a minimimum at $x \simeq 0.261155$ and, for this value, $f(x)=-0.262258$.