$\sin(40^\circ)<\sqrt{\frac{3}7}$ 
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.

My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\sin^2(40^\circ)<3\cos^2(40^\circ)=3(1-\sin^2(40^\circ))$$
$$7\sin^2(40^\circ)<3$$
$$\sin(40^\circ)<\sqrt{\frac{3}7}$$
Is there another way to prove this inequality?

Applying the identity $\sin 3x=3\sin x-4\sin^3 x$ we get
$$\frac{\sqrt 3}2=\sin 120^\circ=3\sin 40^\circ-4\sin^3 40^\circ$$
Now, consider the polynonial
$$P(x)=8x^3-6x+\sqrt 3$$
We know that $\sin40^\circ$ is a root of $P$. Also, we know that $\sin 40^\circ>\sin30^\circ=1/2$. Differentiating $P$, we see that $P$ is increasing in $[1/2,\infty)$. Furthermore,
$$P\left(\sqrt{\frac 37}\right)=\frac{24}{49}\sqrt{21}-\frac67\sqrt{21}+\sqrt 3=\sqrt {21}\left(-\frac{18}{49}+\sqrt{\frac17}\right)\stackrel*>0$$
Since $\sqrt{3/7}>1/2$ and $P(\sqrt{3/7})>P(\sin 40^\circ)$, then the inequality follows.
$*$: Perhaps this inequality needs a proof.
Since $18^2=324<343=7^3$,
$$\frac{18^2}{7^4}<\frac17$$
