# Interesting approximations using trigonometry

Today on Twitter John D. Cook, writing as @AlgebraFact, posted the following tweet:

In radians, sin(11) is very nearly -1.

(It happens to be -0.9999902…)

I thought that was awesome! So, I (@divbyzero) replied that cos(333) is approximately 1. (It is 0.999961…)

Then @michiexile chimed in, pointing out that cos(355) is closer to -1 than cos(333) is to 1. (It is -0.9999999995…)

Finally, I countered with cos(103993), which is 0.9999999998…

So what’s going on here?

All of this comes from the continued fraction for $\pi$,

${\pi =3+\cfrac {1}{7+\cfrac {1}{15+\cfrac {1}{1+\cfrac {1}{292+\cfrac {1}{1+\cdots }}}}}}$.

It is well known that the convergents of a continued fraction (i.e., the fraction you obtain by truncating a continued fraction at some point) are the best rational approximations for the number. The first ten convergents of ${\pi }$ are

${3}$, ${\displaystyle \frac{22}{7}}$, ${\displaystyle \frac{333}{106}}$, ${\displaystyle \frac{355}{113}}$, ${\displaystyle \frac{103993}{33102}}$, ${\displaystyle \frac{104348}{33215}}$, ${\displaystyle \frac{208341}{66317}}$, ${\displaystyle \frac{312689}{99532}}$, ${\displaystyle \frac{833719}{265381}}$, ${\displaystyle \frac{1146408}{364913}}$.

Notice that ${\displaystyle \pi \approx \frac{22}{7}}$, so ${\displaystyle 11\approx \frac{7\pi }{2}}$. Thus ${\displaystyle \sin (11)\approx \sin \big (\frac{7\pi }{2}\big )=1}$. This was John D. Cook’s observation.

Similarly, ${\displaystyle \pi \approx \frac{333}{106}}$, so ${\displaystyle 333\approx 106\pi }$, and ${\cos (333)\approx \cos (106\pi )=1}$. The other two approximations come from the next two convergents.

After the conversation on Twitter, I started playing a little more.

First, I noticed that

${\tan (111)\approx \sqrt {3}}$.

(${\tan (111)-\sqrt {3}\approx -0.011\ldots }$)

This comes from the third convergent. Since ${\displaystyle \pi \approx \frac{333}{106}}$ implies that ${\displaystyle 111\approx 2\pi \cdot 17+\frac{4\pi }{3}}$, ${\displaystyle \tan (111)\approx \tan \big (2\pi \cdot 17+\frac{4\pi }{3}\big )=\tan \big (\frac{4\pi }{3}\big )=\sqrt {3}}$.

Actually, we can do better than that:

$-\tan(69447)\approx \sqrt{3}$.

($-\tan(69447)-\sqrt{3}\approx-0.000011\ldots$)

The approximation, $\displaystyle \pi\approx\frac{208341}{66317}$ implies that $\displaystyle 69447\approx 2\pi\cdot66317+\frac{5\pi}{3}$, so  $\displaystyle \tan(69447)\approx \tan\big(2\pi\cdot66317+\frac{5\pi}{3}\big)=\tan\big(\frac{5\pi}{3}\big)=-\sqrt{3}$.

Next, I noticed that

${\sec (26087)\approx \sqrt {2}}$.

(${\sec (26087)-\sqrt {2}\approx -0.0000039\ldots }$)

This comes from ${\displaystyle \pi \approx \frac{104348}{33215}}$, which implies that ${\displaystyle 26087\approx 2\pi \cdot 4151+\frac{7\pi }{4}}$. Thus ${\displaystyle \sec (26087)\approx \sec \big (2\pi \cdot 4151+\frac{7\pi }{4}\big )=\sec \big (\frac{7\pi }{4}\big )=\sqrt {2}}$

Finally, I noticed that

${-\sec (26087)-\sec (95534)\approx \sqrt {6}}$.

(${-\sec (26087)-\sec (95534)-\sqrt {6}\approx 0.0000046\ldots }$)

In this case, ${\displaystyle \pi \approx \frac{1146408}{364913}}$ gives ${\displaystyle 95534\approx 2\pi \cdot 15204+\frac{17\pi }{12}}$, and ${\displaystyle \sec (95534)\approx \sec \big (2\pi \cdot 15204+\frac{17\pi }{12}\big )=\sec \big (\frac{17\pi }{12}\big )=-\sqrt {6}-\sqrt {2}}$. We obtain the result by substituting the previous approximation for ${\sqrt {2}}$.

Of course, we also have

$\sqrt{6}=\sqrt{2}\sqrt{3}\approx -\sec(26087)\tan(69447)$.

($-\sec(26087)\tan(69447)-\sqrt{6}=0.0000086\ldots$)

1. John says:

That’s great! You really ran with this.

One footnote: Good approximations to pi lead to great approximations to 1 when you take sines. Taylor series says sin(pi/2 + h) is approximately 1 – 0.5 h^2 for small h. So the error in the approximation for pi gets squared.

2. John,
Thanks for pointing that out. I did notice that some approximations were better than others, but hadn’t thought about why.

Dave

3. Nemo says:

Reminds me of my favorite calculator trick.

Type in a bunch of 5’s: 555555, or whatever.

Press “1/x”.

Press “sin”.

Examine the mantissa of the result. Magic!

1. Excellent, I’d never seen that before! Good use of $\sin x\approx x$ and the radians-to-degrees conversion. I’ll have to show that to my students.

4. Walt says:

Found a small typo in your typesetting: the CF of pi is 3; 7, 15, 1, 292 – you forgot the 1 after the 15 in your nested fraction.

1. Wow, good eyes. Thank you for catching that. I updated the post.

5. Awesome.. You guys really doing some good approximation and bringing some fun to it. How I wish Twitter had been there in my school days.

6. Ragesh G R says:

Hi, seems like you have done gross approximations.

Tan 4pi/3 is no way near sec 4pi/3 . Same goes for 5pi/3. Unless i am missing something :)

1. Thanks for catching the typos. They’re fixed now.

7. batbrat says:

I read through the post several times and discussed it with my friends. I am unable to understand why have you approximated tan(111) using sec(111) and tan(69447) using sec(69447).

This is especially puzzling when I consider that computing sec(4*pi/3) in radians yields -2 and tan(4/3 * pi) is 1.7320! They aren’t even close! I’m utterly puzzled.

I’d really appreciate some insight into your thought process.

On an aside, I’d much appreciate it if you deleted my previous comment. I can’t really edit my comment, so I re-posted.

1. Thanks for catching those typos.

1. batbrat says:

No problem. :)