# Calculating Sines without Calculus

One of my better math teachers in high school was John Brumfield. He taught me Algebra I and Precalculus. He was also exceptional at calculation, a skill that got him into in the artillery core in World War II, where he had to calculate gunnery firings in real time. And he had a great sense of humor. But one of the things that stands out most clearly in my memory from one of his classes is something he may have gotten wrong (perhaps deliberately). At that point in the early 1980s scientific calculators existed, but were quite expensive and not yet integrated into the high school mathematics curriculum. We were probably one of the last classes to spend significant time learning about trig tables, and how to interpolate between values. (A surprisingly useful skill, by the way, even if the reason we learned it no longer applies.) I recall that someone in the class asked him how the numbers in the tables were calculated. E.g. how did the author know that the sine of 47.1 degrees was 0.0238875315 and not 0.0238875326 or some other value? And his answer still sticks with me to this day, and I quote it word for word: “Very accurate graphs”.

It wasn’t until I was a sophomore in college that I learned about Taylor Series, and at that point his answer began to bother me. Did he not know about Taylor series? Or did he not want to bother explaining them to us at that point in time? Given the teacher he was, I found it hard to believe that either he deliberately lied to us, or did not know the right answer, so perhaps and at some point in the pre-computer past, graphs were used to make trig tables? I’ve looked but I’ve never been able to find any evidence for that.

Still lately something else occurred to me. Trigonometry predates calculus. It well predates Taylor series. So just how were sines and cosines calculated before anyone knew what a derivative was? This is surprisingly hard to find an answer to, but there is an answer. In fact, it turns out, the Greeks were able to calculate sines and cosines a couple of millennia before calculus. Here’s how.

First define the sine as the ratio of the lengths of the opposite side and the hypotenuse in a right triangle. (You can do it in a unit circle too, which works out the same.) Now consider a triangle with two 45° angles and, for simplicity, sides of length 1:

By the Pythagorean Theorem, the hypotenuse has length \( \sqrt{1^2 + 1^2} = \sqrt{2} \). Therefore the ratio of either the opposite or adjacent side to the hypotenuse; i.e. the sine or a cosine of a 45° angle is \( \frac{1}{\sqrt{2}} \).

It’s a little trickier, but we can also get an exact expression for a 30-60-90 right triangle as follows. Start with an equilateral triangle (i.e. one whose angles are all one third of a straight angle; i.e. 60°:

Now drop a perpendicular from one corner to the adjacent side:

By symmetry it’s easy to see that this divides one 60° angle into two 30° angles, and thus we have two 30-60-90 triangles, with a hypotenuse of length 2 and one side of length 1. The other side has length \( \sqrt{2^2 – 1^2} = \sqrt{3} \), again by the Pythagorean theorem.

Therefore the ratio of the side opposite the 30° angle to the hypotenuse is 1/2, and now we have the sine of 30°. Similarly, the sine of 60° is \( \frac{\sqrt{3}}{2} \). Now we have the sines of three, admittedly special, angles. But how do you get all the others? There are a few such as 36° that can be done with other geometric constructions. (Hint: 36° = 180°/5) but that’s not enough to get the 900 or so values you’d find in a typical trigonometric table.

The trick is now to use other trigonometric identities. In particular we use the half angle and double angle formulas:

\( \sin{\frac{\theta}{2}} = \sqrt{\frac{1 – \cos{\theta}}{2}} \)

\( \sin{ 2\theta } = 2\sin{\theta}\cos{\theta} \)

This can get us all sines that are even divisors of the known angles. E.g. \( \sin{90°} = 2\sin{45°}\cos{45°} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1 \) and \( \sin{15°} = \sqrt{\frac{1 – \cos{30°}}{2}} = \sqrt{\frac{1 – \frac{\sqrt{3}}{2}}{2}} \)

We can continue in this fashion to calculate the sine of 7.5°, the sine of 3.75°, the sine of 1.875°, the sine of .9375°, the sine of .46875°, the sine of 0.234375°, and so on down about the sine of 0.05859375° which is likely small enough for most trigonometric tables. Of course this still leaves a lot of gaps. How do we fill them in? For example, how do you get the sine of 4°? Now we use the trigonometric addition formula:

\( \sin{ \alpha + \beta } = \sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha} \)

Thus, to a reasonable degree of approximation, we can calculate \( \sin{4°} \approx \sin{(3.75° + 0.234375°)} = \sin{3.75°}\cos{0.234375°} + \sin{0.234375°}\cos{3.75°} \)

How good is this approximation? The answer I get after working this out is 0.069484428 whereas if I just ask my calculator for the sine of 4° I get 0.069756474 so it’s accurate to about 3 decimal places. If that’s not good enough, we can add a smaller term or two. There are probably other tricks you could use to improve the accuracy, and reduce the work.

I of course did this with a calculator. I am truly impressed by the mathematicians of the pre-calculus age who did this by hand, and for many more angles than this. Although I’ve glossed over it here they also had to estimate square roots to make this work. By contrast, a Taylor series offers a similar degree of accuracy with only a few terms and no square roots. For instance with *just one term*, the Taylor series gives you the better approximation of 0.0698131701. That’s with no multiplications or additions at all, other than what you need to convert degrees to radians! (The small angle approximation really does work.) With two terms, we get the even better approximation 0.069756460. We can see why nobody has calculated trig tables like this since Brook Taylor invented his eponymous series almost 300 years ago, and the techniques are almost forgotten. Indeed some advanced texts today even define the sine and cosine functions by their Taylor series expansion and then prove that these happen to be the ratios of the sides of the right triangle to the hypotenuse. Personally I find that a very unsatisfying approach. It seems too much like magic, and gives no sense of why this is true. It is comforting to know that pre-calculus mathematicians (including today’s high school students) don’t have to take trigonometric tables on faith or based on the most accurate graphs they can draw. There is such a thing as trigonometry without calculus.

April 2nd, 2015 at 12:02 am

Ever since I started taking trigonometry in school (a long time ago), I found the subject very fascinating, but I always did wonder as to “how” the trig tables were constructed in the first place. it almost seemed to me that we were missing a key part of the story. I never did get an answer from any of my teachers, they always admitted to not knowing, and as I seemed to be the only one in the class who was interested, we always “moved on”. Now, finally, over 50 years later, thanks to the internet, I have finally tracked down a satisfactory answer. Thank you so much for this! Not surprisingly, perhaps, I still seem to be the only one in my circle of friends who shows the slightest interest!

regards,

Dave