BoJone remembers that his first contact with trigonometric functions was around the fifth or sixth grade of elementary school, when I took my cousin’s junior high school mathematics textbook to read. After seeing the chapter on trigonometric functions, I was very interested and hoped to find a method to calculate the values of trigonometric functions based on the angle. However, the book only taught me to use a calculator or look up tables, which greatly disappointed me as a child who loved calculation. This problem was not resolved until my first year of senior high school, when I realized it involved Taylor series in calculus...
I remember that in order to find the trigonometric function values for any angle, I once fitted an approximate formula based on the sine values of 30^\circ, 45^\circ, and 60^\circ: \sin A \approx \sqrt{\frac{A}{60}-1/4}
Where A is in degrees. It was roughly applicable for 25^\circ \sim 60^\circ, and the precision seemed to be around two decimal places. Of course, this result looks very crude today, but it was, after all, my “elementary school work”! I leave it here as a memento.
Later, in junior high school, after learning some senior high school knowledge, I tried to find radical expressions for more angles. Besides 30^\circ, 45^\circ, and 60^\circ, we also have the radical expressions for 18^\circ: \begin{aligned} \sin 18^{\circ} &= \frac{\sqrt{5}-1}{4} \\ \cos 18^{\circ} &= \frac{\sqrt{2\sqrt{5}+10}}{4} \end{aligned}
At the same time, the method to find \cos(A-B) given \cos A and \cos B is: \cos(A-B)=\cos A \cos B+\sqrt{(1-\cos^2 A)(1-\cos^2 B)}
The method to find \cos \frac{A}{2} given \cos A is: \cos\frac{A}{2}=\sqrt{\frac{\cos A+1}{2}}
The method to find \cos \frac{A}{3} given \cos A is: \begin{aligned} \cos A &= 4\left(\cos\frac{A}{3}\right)^3-3\cos\frac{A}{3} \\ \cos \frac{A}{3} &= \frac{1}{2} \left( -\sqrt[3]{\sqrt{\cos^2 A-1}-\cos A} +\sqrt[3]{\sqrt{\cos^2 A-1}+\cos A}\right) \end{aligned}
Thus, 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}
Therefore, it is not difficult to write out \cos 1^\circ. Of course, looking at it now, this can only serve as a kind of existence proof and does not have much practical value.
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