# Inverse sines near a half: Secrets of the Mathematical Ninja

“So, $\sin(x) = 0.53$,” said the student.

“32 degrees,” said the Mathematical Ninja.

The student frowned - the Mathematical Ninja’s showing off was starting to wear her down - and typed it into the calculator to check. “$32.005^º$, *actually*.”

“I’ll take that,” said the Mathematical Ninja.

“How did you guess that?”

“*GUESS?!*” exploded the Mathematical Ninja. “The Mathematical Ninja does not *guess*. The Mathematical Ninja *calculates*.”

“OK, how did you calculate that?”

“Oh, very simply,” said the Mathematical Ninja, brightening up. “There’s a neat thing about inverse sines near a half - it turns out that the difference in degrees from 30 is roughly the percentage difference divided by three.”

“Come again?”

“So, here, 0.53 was 6% more than a half.”

The student thought for a moment. “OK, I buy that.”

“So the degree difference would be 2 degrees?”

“Exactly right.”

“You mean, almost exactly right.” In her head, the student stuck her tongue out.

“So, if I had $\sin(x) = 0.464$, you’d say that’s… 0.036 below, so 7.2%? And the angle difference would be 2.4 *down*…” She thought for a moment. “$27.6º$?”

“$27.645^º$, *actually*,” said the Mathematical Ninja. In their head, they stuck their tongue out.

### Why it works

This involves a little bit of C3 trigonometry, the small-angle approximations $\sin(x) \simeq x$ and $\cos(x) \simeq 1$, when $x$ is in radians, and a neat coincidence.

We’re trying to solve $\sin(x) = 0.5 + e$, where $e$ is relatively small. ((It works fairly well for $|e| < 0.2$.)) Since we know $\sin\left(\frac \pi 6\right) = 0.5,$ ((I wouldn’t normally use 0.5 rather than $\frac 12$, but it fits the context well here.)), we can use that as our starting point:

$\sin\left(\frac \pi 6 + k\right) = 0.5 + e$, where $k$ is a small angle in radians. Let’s expand the compound angle:

$\sin\left(\frac {\pi} {6}\right) \cos(k) + \cos\left(\frac \pi 6\right) \sin(k) = 0.5 + e$

Only, we know $\sin\left(\frac \pi 6\right)=0.5$ and $\cos\left(\frac \pi 6\right) = \frac{\sqrt{3}}{2}$, so we have:

$ 0.5 \cos(k) + \frac{\sqrt{3}}{2} \sin(k) = 0.5 + e$

Because $k$ is small, we’ll say $\cos(k) \simeq 1$ and $\sin(k) \simeq k$:

$ 0.5 + \frac{\sqrt{3}}{2} k \simeq 0.5 + e$, so $\frac{\sqrt{3}}{2}k \simeq e$.

We can work $k$ out from there: it’s clearly $k \simeq e \frac{2}{\sqrt 3}$, in radians - or about 1.154 times as big as $e$.

Converting that into degrees, like kids, we get that the angle change is about 66 times as big as the $e$ - or about two-thirds of $100e$.

So, to get the angle difference, you need to multiply $e$ by 200 (which gives you the percentage difference from 0.5) and then divide by 3. Neat, eh?

You can also go the other way: if you want to know the sine of an angle near 30 degrees, treble the difference, divide by 200 and add on a half. $\sin(28^º)$ is nice and close to 0.47.

* Edited 2021-01-02 to give the Mathematical Ninja the correct gender.

* Edited 2021-06-16 to fix runaway LaTeX