# Exact trig values - Secrets of the Mathematical Ninja

The Mathematical Ninja surreptitiously pressed a button under the table. There was a flash, a sizzle and a slight smell of burning.

The student prodded the on-button of his calculator increasingly frantically.

“Oh dear,” said the Mathematical Ninja. “It must have been a passing electromagnetic storm that’s permanently fried the circuit board of that… *machine*. It’s lucky you didn’t need it for the sum you were doing, eh?”

“But… but… it’s a trig question!” said the student. “You can’t do trig without a calculator, stands to reason!”

“Ha!” said the Mathematical Ninja, and sketched an equilateral triangle on the board ((I say ‘sketched’; any deviations from a perfect equilateral triangle were on the microscopic level and due to imperfections in the pen and/or whiteboard manufacture.)) “What’s that?”

“It’s a triangle.”

“What kind of triangle?”

“An equilateral triangle.”

“So this angle here is…?”

“Sixt… I mean, $\frac{\pi}{3}$.”

The Mathematical Ninja nodded, and bisected the triangle.

“Now it’s right-angled,” said the student, eager to get an answer in before the question. “The angle at the top is $\frac{\pi}{6}$.”

“Correct. How about the side lengths?”

“How big was the triangle?”

“The sides have length $x$,” said the Mathematical Ninja.

The student glowered. “So, the hypotenuse is $x$, the adjacent side is $\frac{x}{2}$ and the remaining side… God… the square root of $x^2 - (x/2)^2$?”

“You can simplify that.”

“I’m sure my calculator could.”

“Unfortunately, your calculator has gone to silicon heaven.”

“FINE. Well, the $x$ can come out… is it $\sqrt{\frac 34}$?”

“It is, but we normally say $\frac{\sqrt{3}}{2}$.”

“Ah, ok. So, $\sin(\pi / 3) = \frac{\sqrt{3}}{2}$ and $\cos(\pi/3) = \frac{1}{2}$. With $\pi/6$, it’s the other way around.”

“Good. You can remember the areas of the squares on this bisected equilateral triangle as 1, 3 and 4, and…”

“I presume I can develop the trig functions from there. Any others I should know?”

“Yes, the isosceles right-angled triangle.” Again, a flurry of sketching.

“So, I could set up the areas of the squares as 1, 1 and 2 - I know the angle is $\frac{\pi}{4}$, so $\sin(\pi/4)$ is the square root of $\frac{1}{2}$.”

“Who needs a calculator, eh?”

“Well… what about angles above $\frac{\pi}{2}$? What then?”

“Another day,” said the Mathematical Ninja, checking his digital watch - which had mysteriously stopped working. “Another day.”