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 his head, he stuck his 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.1 Since we know $\sin\left(\frac \pi 6\right) = 0.5,$2, 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.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. It works fairly well for $|e| < 0.2. []
  2. I wouldn't normally use 0.5 rather than $\frac 12$, but it fits the context well here. []

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