<s>[INST] Proving an inequality of trigonometric functions I am trying to prove that $$(\cos x)^{\cos x} > (\sin x)^{\sin x}$$ for $x \in \left(0, \frac{\pi}{4}\right)$.
I have plotted the graph of $y=(\cos x)^{\cos x} - (\sin x)^{\sin x}$ and see that my conjecture is supported.
Therefore, I tried to prove an equivalent statement $$ \cos x  \ln  \cos x> \sin x \ln \sin x$$.
I tried to use Calculus to prove the statement.
Let $$f(x)= \cos x  \ln  \cos x-\sin x \ln \sin x$$
then $$f'(x)=-\sin x\ln \cos x -\cos x \ln \sin x -\sin x - \cos x.$$
From here I don't know how to proceed.
Any help to prove the inequality will be much appreciated!
 [/INST] (sin x + cos x)^cos x   =  sin x^cos x + cos x^cos x + P1( sin x, cos x)
(sin x + sin x) ^sin x  = sin x^sin x + cos x^sin x +  P2( sin x, cos x)
Where P1 and P2 are polynomials with factors sin x and cos x.
But cos x > sin x in range 0 < x <   π /4 , therefore:
(sin x + cos x)^cos x  >  (sin x + sin x) ^sin x;
⇒ sin x^cos x + cos x^cos x + P1( sin x, cos x) >  sin x^sin x + cos x^sin x 


*

*P2( sin x, cos x)


⇒  cos x^cos x  > sin x^sin x + cos x^sin x - sin x^cos x + P2(sin x, cos x) - P1( sin x, cos x) 
Lim (cos x^sin x - sin x^cos x ) =1 when x    →  0
Lim(cos x^sin x - sin x^cos x ) = 0 when  x   →   π /4
⇒  0 < (cos x^sin x - sin x^cos x ) < 1 in range 0 < x <   π /4 
Lim [ P2( sin x, cos x) - P1( sin x, cos x) ] = 0 when x  →  0
Lim [ P2( sin x, cos x) - P1( sin x, cos x) ] = 0 when x  →   π /4
⇒  [ P2( sin x, cos x) - P1( sin x, cos x) ]   ≈   0 in range  0 < x <  π /4 
Therefore :in range  0 < x <  π /4,  cos x^cos x  >   sin x^sin x 
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