$\sin(40^\circ)<\sqrt{\frac{3}7}$

We need to prove that $$\frac{1-\cos80^{\circ}}{2}<\frac{3}{7}$$ or $$\sin10^{\circ}>\frac{1}{7}.$$

Let $\sin10^{\circ}=x$.

Thus, $$3x-4x^3=\frac{1}{2}$$ or $f(x)=0$, where $$f(x)=x^3-\frac{3}{4}x+\frac{1}{8}$$ and since $$f\left(\frac{1}{7}\right)=\frac{1}{343}-\frac{3}{28}+\frac{1}{8}=\frac{57}{2744}>0,$$ we are done!

Indeed, $f'(x)=3x^2-\frac{3}{4}=3\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)$,

which says that $\sin10^{\circ}$ is an unique root of the equation on $\left(0,\frac{1}{2}\right].$