Prove that $$\frac{1}{3} < \sin{20°} < \frac{7}{20}$$

Attempt $$\sin60°=3\sin20°-4\sin^{3}(20°)$$ Taking $\sin20°$=x

I got the the equation as $$8x^3-6x+\sqrt{3} =0$$ But from here I am not able to do anything. Any suggestions?

Thanks! Edit-graph of p(x) enter image description here

  • 1
    $\begingroup$ Hint: if $p(x)$ is your cubic, what is $p\left( \frac 13\right)$? What is $p\left( \frac 7{20}\right)$? That, plus a quick sketch of the graph should do it. $\endgroup$ – lulu Dec 3 '18 at 14:03
  • $\begingroup$ @lulu In that region p(x) is decreasing so p(1/3)>p(x)>p(7/20) and hence $\frac{1}{3} <x<\frac{7}{20}$ $\endgroup$ – jayant98 Dec 3 '18 at 14:12
  • $\begingroup$ Well, I think this is what you are trying to prove, no? I don't see what argument you are making other than to restate the question. $\endgroup$ – lulu Dec 3 '18 at 14:13
  • $\begingroup$ Sketch the graph! What can you say about the roots of $p(x)$? Can you roughly describe their locations? $\endgroup$ – lulu Dec 3 '18 at 14:13
  • $\begingroup$ Yes this is what I was trying to prove. I wanted affirmation from your side only to be only sure about my steps. But, thanks for help. $\endgroup$ – jayant98 Dec 3 '18 at 14:15

Let $p(x)$ denote the cubic.

Since $$\lim_{x\to - \infty}p(x)=-\infty<0\quad \quad p(0)>0\quad \quad p(.5)<0\quad \quad\lim_{x\to + \infty}p(x)=+\infty$$ we see that one root is negative, a second lies between $0$ and $.5$ and the third is greater than $.5$

Of course $\sin(20^{\circ})$ is positive and, since $0<20<30$ we see that $\sin(20^{\circ})<.5$ Thus the root we want is the middle one of the three. We then remark that $$p\left( \frac 13 \right)>0 \quad \& \quad p\left( \frac 7{20}\right)<0$$ so the root we care about must be between $\frac 13$ and $\frac {7}{20}$ and we are done.


Mostly for fun, here's a way to show that


with only a small amount of multi-digit arithmetic:


and $12^2=144\lt147=3\cdot7^2$, which gives the first inequality, while

$$6\left(7\over20\right)-8\left(7\over20\right)^3={7\over10}\left(3-\left(7\over10\right)^2 \right)={7\over10}\left(300-49\over100 \right)\gt{7\over10}\cdot{250\over100}={7\over4}$$

and $7^2=49\gt48=3\cdot4^2$ gives the second inequality.

  • $\begingroup$ Thanks for another way to attempt such inequalities. $\endgroup$ – jayant98 Dec 3 '18 at 16:43

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