$\sin(40^\circ)<\sqrt{\frac{3}7}$ 
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?
 A: 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 $\left(0,\frac{1}{2}\right].$
A: 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 $18^2=324<343=7^3$,
$$\frac{18^2}{7^4}<\frac17$$
A: Result: $\sin x\geq x-\dfrac{x^2}{2}\dots(*)$

Now $\sin \dfrac{2\pi}{9}=\dfrac{1}{\sqrt2}(\cos\dfrac{5\pi}{180}-\sin\dfrac{5\pi}{180})$
To show $\sin \dfrac{2\pi}{9}<\sqrt{\dfrac{3}{7}}\Leftrightarrow (\cos\dfrac{5\pi}{180}-\sin\dfrac{5\pi}{180})<\sqrt{\dfrac{6}{7}}\Leftrightarrow\sin\dfrac{\pi}{18}>\dfrac{1}{7}$
Using $(*)$, you get $\sin\dfrac{\pi}{18}>\Big(\dfrac{\pi}{18}\Big)-\dfrac{1}{2}\Big(\dfrac{\pi}{18}\Big)^2>\dfrac{1}{7}$
(You can see this calculation here, or you can do by hand by taking $\pi\sim 3.14$)
Hence $$\sin(40^\circ)<\sqrt{\frac{3}7}$$
A: Let $\sin\theta=\sqrt{\frac37}$, so that $\cos\theta=\sqrt{\frac47}$ and $$\sin(\theta-30°)=\sqrt{\frac37}\sqrt{\frac34}-\sqrt{\frac47}\sqrt{\frac14}=\frac1{\sqrt{28}}.$$
Then by the triple angle formula
$$\sin(3\theta-90°)=\frac3{\sqrt{28}}-\frac4{28}\frac1{\sqrt{28}}=\frac{10}{7\sqrt 7}\approx0.54,$$
while 
$$\sin(3\cdot40°-90°)=\frac12.$$

The shift by $90°$ is to stay in a monotonic section of the sinusoid.
A: 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
$$ \cos\left(\frac{4\pi}{9}\right)>\left(\frac{5}{12}\right)^2.$$
