# Why does this nice method work for expressing accurate trigonometric values in the form $\sqrt{\frac{2\pm\sqrt{2\pm\sqrt{2\cdots\pm\sqrt{2}}}}{2}}$?

I am amazed by the nice work of Mr. Daahal on Breaking-Classical-Rules-in-Trigonometry-Exact-Trigonometric-Values. He provided the algorithm without proof. If someone can provide insight on why this works, I would appreciate.

For your convenience, here is a summary of the approach based upon my understanding.

If your target is $$\sin 20^\circ$$, go through the following process :

1. Define $$\frac{45}{2^n}$$ as central values, where n is a natural number.
2. Start from first central value of 45, add or subtract the next central value from previous result to approach the target angle. You can stop when reaching the target, or the precision you need or a repeated pattern.
3. In the following flow chart, always set $$\color{blue}{\boxed{+\sqrt 2}}$$ as fixed starting reference. From then on, compare the sign in present LHS to the sign in previous LHS, if it changes, list $$\color{red}{-\sqrt 2}$$, otherwise, list $$\color{blue}{+\sqrt 2}$$.
4. Gather all the $$\sqrt 2$$'s underneath a single square root sign and divide it by 2, then you have a pretty acurrate value of $$\sin 20^\circ$$.

\begin{align} (0+\frac{45}{2^0}=45>20?\color{blue}{\boxed{+\sqrt 2}})\to(45-\frac{45}{2^1}=22.5>20?\color{red}{-\sqrt 2})\to (22.5-\frac{45}{2^2}=11.25>20? \color{blue}{+\sqrt 2)})\to (11.25+\frac{45}{2^3}=16.875>20?\color{red}{-\sqrt 2)}\to (16.875+\frac{45}{2^4}=19.6875>20?\color{blue}{+\sqrt 2)})\to (19.6875+\frac{45}{2^5}=21.09375>20?\color{blue}{+\sqrt 2})\to (21.09375-\frac{45}{2^6}=20.390625>20?\color{red}{-\sqrt 2})\to (20.390625-\frac{45}{2^7}=19.863125<20?\color{blue}{+\sqrt 2})...\end{align}

Within 0.2% error, \begin{align}\sin 20^\circ= \sqrt{\frac{+2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt 2+\sqrt 2-\sqrt 2+\sqrt 2}}}}{2}} \approx 0.342660717312 \end{align}

• You will see some of the hints in the comments here. – rtybase Apr 15 at 23:23