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

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?

• Without checking carefully, looks OK and a reasonable approach. – Ethan Bolker Aug 21 '17 at 14:05
• "Is there another way to prove this inequality?" Maybe numerical calculation? – Jack Aug 21 '17 at 14:13
• @Jack Numerical calculation, "without the use of a calculator"? Good luck! – Professor Vector Aug 21 '17 at 14:25
• Where did this question come from? – hypergeometric Aug 21 '17 at 14:39
• @hypergeometric I was asked by one of my friend. – Mathxx Aug 22 '17 at 8:25

## 5 Answers

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 $\left(0,\frac{1}{2}\right].$

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

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.

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 $18^2=324<343=7^3$, $$\frac{18^2}{7^4}<\frac17$$

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 $$\cos\left(\frac{4\pi}{9}\right)>\left(\frac{5}{12}\right)^2.$$