Without using a calculator, which is bigger: sin(2°)/2 or sin(3°)/3 ? (Part 2)

In part 1, I proved that sin(2°)/2> sin(3°)/3 using trigonometry.  Here is a proof using power series.

In “Some Simple Lemmas For Bounding Power Series (Part 1)” Example 1, we applied the lemmas to prove that

$$x-x^3/6 + x^5/120>\sin(x) > x-x^3/6$$when $0<x<\sqrt{3}.$  (That’s actually true for all $x>0$, but I did not prove it.)

Let $\epsilon = 1^\circ$. Note that $\epsilon = \frac{\pi}{180} < 1/40.$

Then,

\begin{aligned}\frac{\sin(2^\circ)}2 – \frac{\sin(3^\circ)}3 &= \frac{\sin(2\epsilon)}2 – \frac{\sin(3\epsilon)}3\\&=\frac{ 3\sin(2\epsilon) – 2 \sin(3\epsilon)}6\\&>\frac{ 3 (2\epsilon – (2\epsilon)^3/6) – 2 (3\epsilon – (3\epsilon)^3/6 + (3\epsilon)^5/120 )}6\\&=\frac{ 3(2\epsilon – 4 \epsilon^3/3) – 2 (3\epsilon – 9\epsilon^3/2 + 81\epsilon^5/40 )}6\\&=\frac{ 6\epsilon – 4 \epsilon^3 – (6\epsilon – 9\epsilon^3 + 81\epsilon^5/20 )}6\\&=\frac{ – 4 \epsilon^3 + 9\epsilon^3 – 81\epsilon^5/20 }6\\&=\frac{ 5 – 81\epsilon^2/20 }6 \epsilon^3\\&>\frac{ 5 – 1/20 }6 \epsilon^3 \\&>0 \end{aligned}which proves sin(2°)/2> sin(3°)/3.