The Mathematical Ninja and sin⁡(15º)

The Mathematical Ninja sniffed. “$4\sin (15º)$? Degrees? In my classroom?”

“Uh uh sorry, sensei, I mean $4\sin \left(\frac{\pi }{12}\right)$, obviously, I was just reading from the textmmmff.”

“Don’t eat it all at once. Now, $4\sin \left(\frac{\pi }{12}\right)$ is an interesting one. You know all about Ailes’ Rectangle, of course, so you know that $\sin \left(\frac{\pi }{12}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}$, which makes the whole thing $\sqrt{6}-\sqrt{2}$. Now, obviously, that’s the correct, exact answer. But…”

“Hmmmf. Thank you.”

“… the textbook wants it to three decimal places, for some unfathomable reason. Don’t touch that!”

“Wouldn’t dream of it, sensei!” said the student, fingers rapidly receding from the calculator.

“And it turns out to be 1.035.”

“I imagine you’re going to tell me how you know that.”

“Of course I am. I know it because I can square and square-root things. In particular, $16{\sin }^2\left(\frac{\pi }{12}\right)={(\sqrt{6}-\sqrt{2})}^2=(6-2\sqrt{12}+2)$, or $8-\sqrt{48}$.”

“Isn’t that $8-4\sqrt{3}$?” asked the student, eagerly.

The Mathematical Ninja’s head inclined slightly. “It is indeed. However, the $\sqrt{48}$ is more useful for my purposes; as I’m approximating things, it’s easy to see that $\sqrt{48}\approx 7-\frac{1}{14}$.”

“Ah! So the $16{\sin }^2$ bit is… $1+\frac{1}{14}$, which I could call 1.07?”

“Indeed.”

“Which you can square root to get $1+\frac{1}{28}$, or about 1.035!”

“Actually, $\frac{29}{28}=1.03\dot{5}7142\dot{8}$, but there’s a bit of a finesse in the square root.”

“Can one do the same for $\sin \left(\frac{5}{12}\pi \right)$? I know that’s $\frac{\sqrt{6}+\sqrt{2}}{4}$, so $16{\sin }^2\left(\frac{5}{12}\pi \right)=8+\sqrt{48}$, which is just a shade short of 15.”

“Fourteen and thirteen fourteenths,” muttered the Mathematical Ninja.

“In particular, it’s $\frac{15}{14}$ short of 16, and… I suppose we can take $\frac{15}{14\times 16}$ to be $\frac{1}{15}$, can’t we? So we get ${\sin }^2\left(\frac{5}{12}\pi \right)\approx 1-\frac{1}{15}$, the square root of which is roughly $1-\frac{1}{30}$. So is the answer about $\frac{29}{30}$?”

“Good enough for government work,” said the Mathematical Ninja. “Good enough.”