I have two proofs, both of which aren't very nice. Note that on $(0, 90^{\circ}), \sin{x}$ is increasing. Note also that $\frac{7}{20} < \frac{1}{2} = \sin{30^{\circ}},$ so if $0 > \sin{9x} > -1, 20^{\circ} < x < 30^{\circ}.$ Then, if $\sin{9x} > 0, x < 20^{\circ}$. Applying the triple-angle identity twice, we get: If $\sin{x} = t, \sin{9x} = -4(-4t^3 + 3t)^3 + 3(-4t^3 + 3t).$ If $t = \frac{1}{3}$, we get $\sin{9x} = \frac{1633}{19683}.$ If $t = \frac{7}{20}$, we get $\sin{9x} = -\frac{152945093}{2000000000}.$ Hence the result. Edit: Alternatively, just use the triple-angle identity once, and square; compare the square to $\frac{3}{4}$. At $t = \frac{1}{3}, \sin^2{3x} = \frac{529}{729} < \frac{3}{4}.$ At $t = \frac{7}{20}, \sin^2{3x} = \frac{3087049}{4000000} > \frac{3}{4}.$

We know that $sin3t=3sint-4sin^3t.$ So actually taking $z=sin20$, we have $sin60=3z-4z^3.$ Also note that the function $f(x)=3x-4x^3$ is strictly increasing on $[0,\frac{1}{2}]$. Now the inequality is true for $f(\frac{1}{3})\le \frac{sqrt3}{2} \le f(\frac{7}{20})$.

opptoinfinity wrote: We know that $sin3t=3sint-4sin^3t.$ So actually taking $z=sin20$, we have $sin60=3z-4z^3.$ Also note that the function $f(x)=3x-4x^3$ is strictly increasing on $[0,\frac{1}{2}]$. Now the inequality is true for $f(\frac{1}{3})\le \frac{\sqrt3}{2} \le f(\frac{7}{20})$. edited