3
$\begingroup$

The formula in question:

$$\sin\left(\frac{x}{2^n}\right) = \sqrt{a_1-\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\dots+\sqrt{a_{n-1}+\sqrt{\frac{a_{n-1}}{2}\left(1-\sin^2(x)\right)}}}}}}$$ where $$a_k = \frac{1}{2^{2^k-1}} \quad \forall k \in \{1,2,\dots,n-1\}, \ n \in \Bbb{N}, \ x \in \left[0,\frac{\pi}{2}\right[$$ and only the first sign (after $a_1$) is $-$, the rest is $+$.

If this holds, this is a great way to calculate the $\sin$ of small angles, specifically ones that are a power of $\frac{1}{2}$ radians.

My attempt:

I have derived at this by continously using the formula $$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{1-\sqrt{1-\sin^2(x)}}{2}} = \sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{\sin^2(x)}{4}}}$$ Which holds true, since $$ \begin{align} \sin(2x) &= 2\sin(x)\cos(x) \\ \sin(x) &= 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) \\ \sin(x) &= 2\sin\left(\frac{x}{2}\right)\sqrt{1-\sin^2\left(\frac{x}{2}\right)} \\ \sin^2(x) &= 4\sin^2\left(\frac{x}{2}\right)\left(1-\sin^2\left(\frac{x}{2}\right)\right) \ \left(\text{if } x \in \left[0,\frac{\pi}{2}\right[\right) \\ \sin^2(x) &= 4\sin^2\left(\frac{x}{2}\right)-4\sin^4\left(\frac{x}{2}\right) \\ 0 &= 4\sin^4\left(\frac{x}{2}\right)-4\sin^2\left(\frac{x}{2}\right)+\sin^2(x) \\ \sin^2\left(\frac{x}{2}\right)_{1,2} &= \frac{4 \pm \sqrt{16-16\sin^2(x)}}{8} = \frac{1 \pm \sqrt{1-\sin^2(x)}}{2} \end{align} $$

And this holds true with $-$, since: $$ \begin{align} \sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} \frac{1-\sqrt{1-\sin^2(x)}}{2} \\ 2\sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} 1-\cos(x) \\ 2\sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} 1-\cos^2\left(\frac{x}{2}\right)+\sin^2\left(\frac{x}{2}\right) \\ \sin^2\left(\frac{x}{2}\right) &\stackrel{?}{=} 1-\cos^2\left(\frac{x}{2}\right) \end{align} $$

Yes, since $$ \sin^2\left(\frac{x}{2}\right) + \cos^2\left(\frac{x}{2}\right) = 1 $$ So we found that if $x \in \left[0,\frac{\pi}{2}\right[$, then $$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{1-\sqrt{1-\sin^2(x)}}{2}}$$ Now I plugged the formula into itself a couple times, and guessed what it would look like if I had plugged it in $n$ times. However, I have only assumed the above values of $a_k$ are true by looking at the results, so I'd like a rigorous proof of the formula.

I tried induction by $n$, but I couldn't figure out the $n \rightarrow n+1$ step.

Question:

Provide a proof of the first formula or correct it if it's wrong.

$\endgroup$
7
  • 1
    $\begingroup$ Use $\sin\left(\frac{x}{2}\right) = \sqrt{\frac{1-\sqrt{1-\sin^2(x)}}{2}}$. Perform $x \rightarrow x/2$ and use the above relation again. Performing these operations repeatedly. $\endgroup$ – user600016 Oct 9 '19 at 15:33
  • $\begingroup$ I've done that, however the calculations really get messy, and I'm not certain about the result that I got. Especially the $a_k$ values. $\endgroup$ – Daniel P Oct 9 '19 at 15:35
  • $\begingroup$ Are you looking for some analytical verification of the formula you proposed? If yes, you could run a simple code in some programming language. $\endgroup$ – user600016 Oct 9 '19 at 15:37
  • 2
    $\begingroup$ If this holds, this is a great way to calculate the sin of small angles. For this, use Taylor series expansion instead. $\endgroup$ – Jean-Claude Arbaut Oct 9 '19 at 15:41
  • 2
    $\begingroup$ I disagree. The Taylor expansion is an infinite sum, this is a finite method. And since we only have to plug the formula into itself $n$ times, this is a fast and computable method. $\endgroup$ – Daniel P Oct 9 '19 at 15:42
1
$\begingroup$

It's a bit tedious, but nonetheless doable:

Since we have

\begin{align}\sin\left(\frac{x}{{2^{n+1}}}\right)&= \sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{\sin^2\left(\frac{x}{{2^{n}}}\right)}{4}}} \end{align}

We can use the induction hypothesis to calculate \begin{align*}\frac{1}{4}\cdot\sin^2\left(\frac{x}{{2^{n}}}\right)&=\frac{1}{4}\cdot \left(a_1-\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\dots+\sqrt{a_{n-1}+\sqrt{\frac{a_{n-1}}{2}\left(1-\sin^2(x)\right)}}}}}\right)\\ &=\frac{a_1}{2^2}-\sqrt{\frac{a_2}{(2^{2})^2}+\sqrt{\frac{a_3}{((2^{2})^2)^2}+\sqrt{\frac{a_4}{2^{2^4}}+\dots+\sqrt{\frac{a_{n-1}}{2^{2^{n-1}}}+\sqrt{\frac{a_{n}}{2^{2^n}\cdot 2}\left(1-\sin^2(x)\right)}}}}}\\ &=a_2-\sqrt{a_3+\sqrt{a_4+\sqrt{a_5+\dots+\sqrt{a_{n}+\sqrt{\frac{a_{n+1}}{2}\left(1-\sin^2(x)\right)}}}}} \end{align*} where we used that $$\frac{a_n}{2^{2^n}}=\frac{1}{2^{2^n-1}\cdot 2^{2^n}}=\frac{1}{2^{2\cdot 2^{n}-1}}=\frac{1}{2^{2^{n+1}-1}}=a_{n+1}$$ And since $$\color{red}{a_1=\frac{1}{2^{2^1-1}}=\frac{1}{2}}$$ and $a_2=\frac{1}{2^{2^2-1}}=\frac{1}{8}$ so that $$\color{blue}{\frac{1}{4}-a_2=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}=a_2}$$ you get

\begin{align}\sin\left(\frac{x}{{2^{n+1}}}\right)&= \sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{\sin^2\left(\frac{x}{{2^{n}}}\right)}{4}}}\\ &= \sqrt{\color{red}{\frac{1}{2}}-\sqrt{\color{blue}{\frac{1}{4}-a_2}+\sqrt{a_3+\sqrt{a_4+\sqrt{a_5+\dots+\sqrt{a_{n}+\sqrt{\frac{a_{n+1}}{2}\left(1-\sin^2(x)\right)}}}}}}}\\ &= \sqrt{\color{red}{a_1}-\sqrt{\color{blue}{a_2}+\sqrt{a_3+\sqrt{a_4+\sqrt{a_5+\dots+\sqrt{a_{n}+\sqrt{\frac{a_{n+1}}{2}\left(1-\sin^2(x)\right)}}}}}}} \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.