It took the Mathematical Ninja a little longer than normal; the student had managed to rummage around in her bag and lay a finger on the calculator before simultaneously feeling her arm pulled away by a lasso and hearing “0.3805. Or, as a one-off, since the question is asking for degrees, $\arctan(0.4)\approx 21.8º$.”

The student rubbed her arm and sighed. “Go on, then.”

“As everyone knows, $\tan\br{\piby 8} = \sqrt{2}-1$.”

A roll of the eyes.

“OK, there’s your homework. There’s also the nice result that $\tan\br{\frac{a}{b} + \frac{a-b}{a+b}} = \tan \br{\piby 4}$.”

“Is it?”

“There’s exercise 2 for you.”

The student sighed. “So, $\tan\br{\frac{2}{5} + \frac{3}{7}}=1$. I don’t really see how that helps. $\frac{3}{7}$ is what, 0.42, and that’s not any easier to manipulate than the other one.”

“True, true. So instead, let $\theta = \arctan\br{\frac{2}{5}}$, so that $\tan\br{2\theta} = \frac{20}{21}$.”

“Oh! And that is close to 1! It’s only $\frac{1}{41}$ away, so $2\theta \approx \piby 4 - \frac{1}{41}$ - assuming$\frac{1}{41}$ is small enough that $\tan(x) \approx x$.”

The Mathematical Ninja picked up the lasso and furled it. There would be no further use for it in this class.

“So that makes $\theta \approx \piby 8 - \frac{1}{82}$. That would be around $\frac{31}{80}-\frac{1}{82}$, a bit more than three-eighths, so you’d call it 0.38.”


“And in degrees…” – the student took care to pull a slightly disgusted face – “that’s 22.5 minus a bit. What’s $\frac{57.3}{82}$, roughly? Call it $\frac{19}{27}$, a bit more than two-thirds of a degree. 21.8 degrees.”

A subtle nod. The Mathematical Ninja made a note to give the Mathematical Cowboy the lasso back.