The Radical Expression for cos 1°

cos 1°的根式表达式 苏剑林 · 2011-06-26 · Mathematics function trigonometric function

BoJone remembers that his first encounter with trigonometric functions was probably around fifth or sixth grade, when I borrowed my cousin's middle-school math textbook to read. Upon reaching the chapter on trigonometric functions, I got really into it, hoping to find a way to compute trigonometric values directly from angles. But the textbook only taught me to use a calculator and look up tables, which was a huge letdown for a kid who loved computing things by hand. This problem wasn't actually resolved until my first year of high school, when it turned out to involve the Taylor series from calculus...

I remember that, in order to compute the trigonometric values of arbitrary angles, I once fitted an approximate formula based on the sine values of 30°, 45°, and 60°:

$$\sin A \approx \sqrt{\frac{A}{60}-1/4}$$

where A is in degrees, and it's roughly valid for 25° to 60°, with an accuracy of about two decimal places. Of course, by today's standards this result is quite crude, but after all, it was my "elementary school masterpiece"! I'm keeping it here as a memento.

Later, in middle school, having picked up some high-school-level knowledge, I tried to find trigonometric expressions for more angles. Besides 30°, 45°, and 60°, we also have the trigonometric expression for 18°:

$$\begin{aligned}\sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \\ \cos 18^{\circ}=\frac{\sqrt{2\sqrt{5}+10}}{4}\end{aligned}$$

Also, using $cos A$ and $cos B$ to find $cos (A-B)$:

$$\cos(A-B)=\cos A \cos B+\sqrt{(1-\cos^2 A)(1-\cos^2 B)}$$

Using $cos A$ to find $cos \frac{A}{2}$:

$$\cos\frac{A}{2}=\sqrt{\frac{\cos A+1}{2}}$$

Using $cos A$ to find $cos \frac{A}{3}$:

$$\begin{aligned}\cos A=4(\cos\frac{A}{3})^3-3\cos\frac{A}{3} \\ \cos \frac{A}{3} = \frac{1}{2} ( -\sqrt[3]{\sqrt{\cos^2 A-1}-\cos A} +\sqrt[3]{\sqrt{\cos^2 A-1}+\cos A})\end{aligned}$$

So we can write: $cos 15^{\circ}=\frac{\sqrt{\sqrt{3}+2}}{2}$

$$\begin{aligned}\cos 3^{\circ}=\cos(18^{\circ}-15^{\circ})= \\ \frac{\sqrt{2\sqrt{15}+4\sqrt{5}+10\sqrt{3}+20}+\sqrt{2\sqrt{15}-4\sqrt{5}-6\sqrt{3}+12}}{8}\end{aligned}$$

From there, it's not hard to write down cos 1°. Of course, looking back on it now, this can only serve as a kind of existence proof — it doesn't have much practical value.