Find the greatest integer less than $\frac{1}{\sin^2(\sin(1))}$ without calculator.

Question

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?

Answer 1

We'll prove that $$\frac{1}{\sin^2\sin1}<2,$$ for which it's enough to prove that $$\sin\sin1>\frac{1}{\sqrt2}$$ or $$\sin1>\frac{\pi}{4},$$ for which it's enough to prove that $$\sin54^{\circ}>\frac{\pi}{4}$$ or $$\frac{\sqrt5+1}{4}>\frac{\pi}{4}$$ or $$\sqrt5+1>\pi.$$ Can you end it now?

I got the answer: $1$.

Answer 2

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

so $$1 < \frac{1}{\sin^2(\sin(1))} < 2$$