# A proof for : Let $0<x\leq\frac{\pi}{4}$ then $\sin(x)^{\sin(x)}\leq \cos(x)^{\cos(x)}$

Let $$0 then we have : $$\sin(x)^{\sin(x)}\leq \cos(x)^{\cos(x)}$$

I found that beautiful so I propose a proof without derivatives :

I recall that (for :$$0):

$$\cos(x)=\frac{1}{\sqrt{1+y^2}} \quad \sin(x)=\frac{y}{\sqrt{1+y^2}}$$

Where $$y=\tan(x)$$

Now we make the subsitution $$y=\sinh(a)$$ we have :

$$\cos(x)=\frac{1}{\cosh(a)} \quad \sin(x)=\tanh(a)$$

So we have :

Let $$0 then we have : $$\tanh(a)^{\tanh(a)}\leq \Big(\frac{1}{\cosh(a)}\Big)^{\frac{1}{\cosh(a)}}$$

Now we use the Bernoulli's inequality to get : $$\tanh(a)^{\tanh(a)}\leq (1+\sinh(a)(\tanh(a)-1))^{\frac{1}{\cosh(a)}}$$

Recalling that (See comment of MartinR): $$\tanh(a)^{\tanh(a)}=\Big(1+\tanh(a)-1\Big)^{\tanh(a)}=\Big(1+\tanh(a)-1\Big)^{\frac{\sinh{x}}{\cosh(x)}}$$

So it sufficient to prove :

$$(1+\sinh(a)(\tanh(a)-1))^{\frac{1}{\cosh(a)}}\leq \Big(\frac{1}{\cosh(a)}\Big)^{\frac{1}{\cosh(a)}}$$

Or :

$$(1+\sinh(a)(\tanh(a)-1))\leq \Big(\frac{1}{\cosh(a)}\Big)$$

Multiplying by $$\cosh(a)$$ we get :

$$(\cosh(a)+\sinh(a)(\sinh(a)-\cosh(a)))\leq 1$$

Or:

$$(\cosh(a)+\sinh(a)(\sinh(a)-\cosh(a)))-1\leq 0$$

Putting $$u=e^a$$ and multiplying by $$u^2$$ we get a cubic polynomials wich can be easily factored since we have one as root .

Done!

My question :

Is it right ?

Have you an alternative proof ?