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

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: $1$.


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

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

Tags:

Inequality

Trigonometry

Number Comparison

Ceiling And Floor Functions

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