If $0<x<\frac{\pi}{4}$ then prove that $\left(\sin x\right)^{\sin x}<\left(\cos x\right)^{\cos x}$.

My attempt:

If $0<x<\frac{\pi}{4}$

then $\sin x <\cos x$

which means $\left(\sin x\right)^{\sin x}<\left(\cos x\right)^{\sin x}<\left(\cos x\right)^{\cos x}$.

Am I correct. Can a more rigorous proof be given

  • $\begingroup$ Since $0<\cos x<1$ in the given interval, from $\sin x<\cos x$ you can deduce $(\cos x)^{\sin x}\color{red}{>}(\cos x)^{\cos x}$; not the same inequality as you're using. $\endgroup$
    – egreg
    Jan 2, 2017 at 14:04

1 Answer 1


I saw this solution in a textbook($103$ Problems in Trigonometry by Titu Andreescu) but i wanted a second opinion. The way the solution is it can only be answered by someone who has created the question and the solution.I will quote Titu:

Observe that the logarithmic function is concave down.

We apply Jensen's inequality to the numbers $\sin x<\cos x<\sin x+\cos x$

with weights $\lambda_{1}=\tan x$ and $\lambda_{2}=1-\tan x$(because for $0<x<\frac{\pi}{4}$ we have $\lambda_{1}=\tan x>0$ and $\lambda_{2}=1-\tan x>0$) to obtain

$\log(\cos x)$

$=\log\left(\tan x\sin x+(1-\tan x)(\sin x+\cos x)\right)$

$>\tan x\log(\sin x)+(1-\tan x)\log(\sin x+\cos x)$

Since $\sin x+\cos x=\sqrt{2}\sin(x+\frac{\pi}{4})>1$ and $\tan x<1$

We can say that $\log(\cos x)>\tan x\log(\sin x)$

Multiplying both sides by $\cos x$ and then exponientating we obtain the required result.

Please suggest some other way.


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