The Mathematical Ninja and the Tangents Near 1

“Forty-two degrees,” said the Mathematical Ninja, as smugly as possible while still using degrees.

The student’s hand had barely twitched towards the calculator.

“Go ahead, punk,” said the Mathematical Ninja. “Make my day.”

“Righto,” said the student, and tapped in $\tan^{-1} \left( 0.9 \right)$, carefully closing the bracket. “41.987. That’s not bad, sensei. Not bad at all. I suppose…”

“Of course I’m going to tell you how I did it. There’s an identity for finding the tangent of composite angles: $\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$.”

“I’m not really any the wiser… you’re right, I’ll shut up now.”

“It so happens that $\tan(45º) = 1$… no, there’s no need to check it. That means that for angles near 45º” — the Mathematical Ninja pulled a face, wanting desperately to say $\frac{\pi}{4}$, but knowing the student would be even more lost — “you have $\tan(45 + B) = \frac{1 + \tan(B)}{1 - \tan(B)}$.”

“I’m happy to accept that,” said the student, carefully. “And in this case, because 0.9 is pretty close to 1, I suppose you can do something clever with it?”

“But of course! I can rearrange to find $\tan(B)$!”

The student thought for a second. “Can I try that myself, on paper?”

The Mathematical Ninja thought for a moment, realised that that was the kind of thing that was supposed to be encouraged, and shrugged.

“I reckon, for this one, $0.9 = \frac{1 + t}{1-t}$ — I’ve called $\tan(B)$ “$t$”, I hope that’s ok.”

The Mathematical Ninja nodded. It was a good trick.

“So, $0.9(1-t) = 1 + t$, and…” — pause — "$-0.1 = 1.9t$. That means $\tan(B) = -\frac{1}{19}$.”

“Good so far,” said the Mathematical Ninja.

“I don’t know how to do $\tan^{-1}\left(-\frac{1}{19}\right)$, though,” said the student. “Is there a trick for that?”

The Mathematical Ninja smiled. “Indeed. For small angles — in radians, at least, like the grown-ups use — $\tan(x) \approx x$.”

“So you just need to convert $-\frac{1}{19}$ into degrees?”

“Indeed,” said the Mathematical Ninja. “And because a radian is about 57º, $-\frac{1}{19}$ radians is about -3º.”

“Which you take off to get 42º. Pretty!”

The Mathematical Ninja grinned an enormous grin; the day was well and truly made.


In general, the Mathematical Ninja notes, $\tan^{-1}\left(\frac AB\right) = \frac{\pi}{4} + \tan^{-1}\left(\frac{A-B}{A+B}\right)$, and for $\frac{A}{B} \approx 1$, the second term on the right hand side is roughly $\frac{A-B}{A+B}$.

* Edited 2016-08-31 to fix a typo - thanks, @FennekLyra!

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.

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