“$\cos^{-1}(0.93333)$, said the student. A GCSE student, struggling a little; the Mathematical Ninja bit his tongue rather than correct him to $\arccos$ or to $\frac {14}{15}$; he also accepted, grudgingly, the answer was going to be in degrees.

“21.04”, said the student. “Not too terrible.”

“I suppose you’re wondering,” said the Mathematical Ninja, directly to camera, “how I did that. Well, let me tell you. It went down like this:

* Work out $\sqrt{1-x}$. In this case, that’s the square root of $\frac1{15}$, which is a little less than 0.26.

* Multiply that by 10 to get 2.6

* Work out $9-x$, which is $8 + \frac{1}{15}$

* Multiply the two answers together to get 20.8 plus whatever a fifteenth of 2.6 is - 0.17 or so - to get 20.97.

This is a little on the low side, but still within a tenth of a degree - partly because we got lucky with the rounding (if we did it perfectly on the calculator, we’d get 20.83, which is almost the worst case - this is always within about a quarter of a degree).

### Why does it work?

It comes from the Maclaurin series for $\cos(x)$:

$\cos(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - …$, although the Mathematical Ninja only uses the first couple of terms. If he knows $\cos(x) = c$, he can say:

$c \simeq 1 - \frac 12 x^2$ $x^2 \simeq 2 - 2c$, working in radians ((as God intended)).

That gives $x \simeq \sqrt{2} \sqrt{1-c}$. If you convert $x$ to degrees, you get $81\sqrt{1-c}$, which is a good approximation for smallish $c$, but is off by 9 for $c = 1$. (Mortals like me use 80, which is off by 10, of course.)

However - with a little help from the Mathematical Pirate, we can adjust it to bring it back into line. If you look at the proportional error for the approximation, it’s pretty much a straight line, with equation $\frac{act}{est} \simeq \frac{9}{8} - \frac{c}{8}$. That means we now have: $x \simeq 80\sqrt{1-c} (\frac{9-c}{8})$, or (more neatly): $x \simeq 10 \sqrt{1-c}(9-c)$.

The worst-case answers for this are around c = 0.25 (it’s a quarter-degree too low) and c = 0.883 (a quarter-degree too high).

It’s (for mortals, at least), very difficult to do a Taylor series for arccosine around $x=1$ because the gradient there is infinite. Doing it around a different point almost always leads to a $\pi$ knocking around, and who wants that?

### One last word from the Ninja

“It’s not $10\sqrt{1-x} (9-x)$ if you do it properly,” he says; “It’s more like $\frac{7}{40}\sqrt{1-x} (9-x)$.” He stomps, turns, and mutters something about “bad degrees.”