Proving $(\sin x)^{\sin x}<\cos x$ for $0<x\leq\frac{2\pi}{9}$

Question

Prove that $$(\sin x)^{\sin x}<\cos x$$ for $0<x\leq\frac{2\pi}{9}$.

My try was to do $\sin x\log\sin x-\log\cos x<0$ So I did $f(x)=\sin x\log\sin x-\log\cos x$ and $f'(x)=\cos x\log\sin x+\cos x+\cot x$, but I don't know what to do now.

Answer 1

It suffices to show $F(x)=\log(\cos(x))-\log(\sin(x)^{\sin(x)})$ is non-negative on the given interval $I$. Note $F(2\pi/9)\approx 0.0175591>0$ and we can continuously extend $F$ from the right at $0$ to $F(0)=0$. To show $F$ is non-negative, we'll show it is concave down. We have $$ F''(x)=-\sec ^2(x)-\cos (x) \cot (x)+\sin (x) (1+\log (\sin (x))) $$Now we use the fact that $\sin(x)\log(\sin(x))<0$ on $I$. Then $$ F''(x)<-\sec ^2(x)-\cos (x) \cot (x)+\sin (x)\cdot 1 $$We can multiply this by $\sin(x)$ without changing the sign: $$ (-\sec ^2(x)-\cos (x) \cot (x)+\sin (x))\cdot\sin(x) = -\tan (x) \sec (x)-\cos (2 x)<0 $$Then $F''(x)<0$ on $I$, so $F(x)>0$ on $I$.

Answer 2

This is not an answer; just made for the fun of it !

Considering the function $$f(x)=\log (\cos (x))-\sin (x) \log (\sin (x))$$ expand it composing Taylor series built at $x=0$ and consider the truncet $$f(x)=-x \log (x)-\frac{x^2}{2}+O\left(x^3\right)$$ Consider the function $$g(x)=-x \log (x)-\frac{x^2}{2}$$

The maximum of $g(x)$ is attained at $$x_*=W\left(\frac{1}{e}\right)\approx 0.278465 $$ ($W(.)$ being Lambert function) while an "exact" calculation for $f(x)$ reveals a maximum $x\approx 0.277208$ $$g(x_*) =\frac{1}{2} W\left(\frac{1}{e}\right) \left(2+W\left(\frac{1}{e}\right)\right)\approx 0.317236$$ while the maximum value of $f(x)$ is $\approx 0.315706$.

Similarly, the solution of $g(x)=0$ corresponds to $$x=2 W\left(\frac{1}{2}\right)\approx 0.703467$$ which is not far from $0.711828$ solution of $f(x)=0$ (notice that these nmbers are $ \gt \frac{2\pi}9=0.698132$