Find the greatest integer less than $\frac{1}{\sin^2(\sin(1))}$ without calculator. Find the greatest integer less than $$\frac{1}{\sin^2(\sin(1))}$$
This was on one of my tests. All angles in radians. Here's my work:
$$0<1<\frac{\pi}{3}<\frac{\pi}{2}$$
Since $\sin(x)$ is increasing in the first quadrant,
$$0<\sin(1)<\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}<\sin\left(\frac{\pi}{2}\right)=1<\frac{\pi}{2}$$
$\sin(1)$ radians and $\frac{\sqrt{3}}{2}$ radians also lie in the first quadrant, and $\sin^2(x)$ is increasing in the first quadrant
$$\sin^2(\sin(1))<\sin^2\left(\frac{\sqrt{3}}{2}\right)<\frac{3}{4}$$
$$\frac{1}{\sin^2(\sin(1))}>\frac{4}{3}$$
But I cannot find an upper bound on my expression. Any help?

$$1 >  \frac{7}{24}\pi \approx 0.916$$
$$\sin(1) > \sin\left(\frac{7}{24}\pi\right) = \frac12\sqrt{2+\sqrt{2-\sqrt{3}}} \approx 0.793 > \frac{\pi}{4} \approx 0.785$$
$$\frac{1}{\sin^2(\sin(1))} < \frac{1}{\sin^2(\frac{\pi}{4})}=2$$
so $$1 < \frac{1}{\sin^2(\sin(1))} < 2$$
