As the student was wont to do, he idly muttered “So, that’s $\cos(10º)$…”

The calculator, as calculators are wont to do when the Mathematical Ninja is around, suddenly went up in smoke. “0.985,” with a heavy implication of ‘you don’t need a calculator for that’.

As the student was wont to do, he stifled the urge to say ‘but… we mortals kind of do need a calculator for that.’ Instead, he sighed (as he was wont to do) and let the Mathematical Ninja talk it through.

“In radians, $\cos(x)$ is approximately $1-\frac12 x^2$.”

The student shrugged. He wasn’t doing STEP and didn’t much care for small angle approximations.

“Meanwhile, the conversion factor for degrees to radians - the only way one should ever consider converting - is about $\frac{7}{400}$. That squared is $\frac{49}{160,000}$, or about $\frac{1}{3,200}$, and half of it is $\frac{1}{6,400}$.”

“If you say so, sensei.”

“So, to find $\cos(yº)$, you simply work out $1 - \frac{1}{6,400}y^2$.”

“Simply.”

“Simply. For $y=10$, that’s $1-\frac{1}{64}$, and $\frac{1}{64}$ is about 0.015. (You can get the same result by dividing by 80 and squaring the result).”

“Which you presumably took away from 1 to get 0.985?”

“That, young student, is precisely what I did.”

“What about $\arccos$? Is there a trick for that?”

The Mathematical Ninja’s eyes narrowed slightly, recognising that the only-way-to-convert statement was about to be disproved by counterexample. “Well, of course. At least when $c$ is near 1, $\arccos(1-c)$ can be approximated quite easily, just working backwards. $1-c \approx 1 - \frac{1}{6,400}y^2$, so $y \approx 80\sqrt{c}$.”

“So, $\arccos(0.99)$ would be 0.01 less than 1, square rooted is 0.1, and multiplied by 80 gives 8º?”

“8.1º,” giving a slight wince, as the Mathematical Ninja was wont to do.

Uncharacteristically, the Mathematical Ninja missed a trick when explaining this (MY calculator is fine, thanks for asking): the conversion factor is $\frac{\pi}{180}$, the square of which is closer to $\frac{1}{3280}$; half of that is $\frac{1}{6560}$, which is a razor-fine knife edge away from $\left(\frac{1}{81}\right)^2$. Which is, presumably, where the final, more accurate, approximation comes from.