<s>[INST] $\sin(40^\circ)<\sqrt{\frac{3}7}$ 
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?
 [/INST] The given inequality is equivalent to $\cos^2\left(\frac{2\pi}{9}\right)>\frac{4}{7}$, or to $\cos\left(\frac{4\pi}{9}\right)>\frac{1}{7}$.
The minimal polynomial of $\alpha=\cos\left(\frac{4\pi}{9}\right)$ over $\mathbb{Q}$ can be easily derived from $\Phi_9(x)=x^6+x^3+1$, and it is given by $p(x)=1-6x+8x^3$. Since $p'(x)=-6(1-4x^2)$, $p(x)$ is a decreasing function on $\left[0,\frac{1}{7}\right]$, hence the claim simply follows from checking that $p\left(\frac{1}{7}\right)>0$. A sharper inequality which can be proved through the same technique is
$$ \cos\left(\frac{4\pi}{9}\right)>\left(\frac{5}{12}\right)^2.$$
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