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)
Best Answer
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.