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Sines, Cosines, and infinitely nested radicals

DreamWeaver

Well-known member
Sep 16, 2013
337
The aims of this tutorial are threefold:

(1) By assuming two values of the Cosine function - which will be proven later on - we develop a large number of multiply-nested radicals to express \(\displaystyle \cos(\pi/2^kn),\, \sin(\pi/2^kn), \,\) and \(\displaystyle \tan(\pi/2^kn) \) in closed form, for ever smaller arguments \(\displaystyle (k, n \in \mathbb{Z}^{+})\).

(2) By considering the limiting values of \(\displaystyle \cos(\pi/2^kn),\, \sin(\pi/2^kn), \,\) and \(\displaystyle \tan(\pi/2^kn) \) as \(\displaystyle k\to \infty\), we evaluate a number of infinitely-nested radicals, such as:


\(\displaystyle \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2\)


(3) Finally, we consider three new (???) trigonometric functions, which I will henceforth refer to as the Fractional Sine, Cosine, and Tangent functions respectively:


\(\displaystyle \mathscr{Fs}(\theta) = \sum_{k=0}^{\infty} \sin\left( \frac{\theta}{2^k} \right)\)


\(\displaystyle \mathscr{Fc}(\theta) = \sum_{k=0}^{\infty} (-1)^k\cos\left( \frac{\theta}{2^k} \right)\)


\(\displaystyle \mathscr{Ft}(\theta) = \sum_{k=0}^{\infty} \tan\left( \frac{\theta}{2^k} \right)\)


Naturally, some restrictions will need to be applied to \(\displaystyle \theta\) in each case, to ensure convergence. Also, note that the Fractional Cosine function is an alternating series; this is due to the fact that \(\displaystyle \lim_{\theta \to 0}\, \cos (\theta) = 1\).



--------------------------------------------------




For now, we will assume the following two values of the Cosine:


\(\displaystyle (01) \quad \cos\left(\frac{\pi}{2}\right) = 0\)


\(\displaystyle (02) \quad \cos\left(\frac{\pi}{5}\right) = \frac{1}{2} \sqrt{ \frac{3+\sqrt{5} }{2} }\)


In addition, the trigonometric identities used herein are elementary, so proofs will be omitted. The main identities used, aside from \(\displaystyle \cos^2 x+ \sin^2 x = 1\), are essentially three different forms of the Cosine Double Angle formula:


\(\displaystyle (03)\quad \sin\frac{x}{2} = \pm \sqrt{\frac{1-\cos x}{2}}\)


\(\displaystyle (04)\quad \cos\frac{x}{2} = \pm \sqrt{\frac{1+\cos x}{2}}\)


\(\displaystyle (05)\quad \tan^2x= \frac{1-\cos 2x}{1+\cos 2x}\)



In identity (03), we have \(\displaystyle \text{sgn}= + \Leftrightarrow 0 < x < 2\pi\), whereas in identity (04) we have \(\displaystyle \text{sgn}= + \Leftrightarrow -\pi < x < \pi\).



--------------------------------------------------




Entering \(\displaystyle \cos(\pi/2)= 0\) in the RHS of identity (04) gives:


\(\displaystyle \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\)


If we repeat this process, each time entering our new value in the RHS of identity (04), we obtain the series of values:


\(\displaystyle \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)


\(\displaystyle \cos\left(\frac{\pi}{8}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{2} }\)


\(\displaystyle \cos\left(\frac{\pi}{16}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }\)


\(\displaystyle \cos\left(\frac{\pi}{32}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }\)


\(\displaystyle \cos\left(\frac{\pi}{64}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } }\)


\(\displaystyle \cos\left(\frac{\pi}{128}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } } }\)



We introduce the operator \(\displaystyle \Delta_n(x)\), where the operation is "add 2 and then square-root the sum" \(\displaystyle n\)-times. Here, \(\displaystyle x\) is the operand, that is the initial value that is to be operated on. Then we man write:


\(\displaystyle \cos\left(\frac{\pi}{2^{n+1}}\right) = \frac{1}{2}\, \Delta_n(0)\quad \quad \quad n=0, 1, 2, \cdots \)


Now, on the one hand, we have


\(\displaystyle \text{limit}_{n\to \infty} \, \Delta_n(0) = \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } }\)


whereas on the other,


\(\displaystyle \text{limit}_{n\to \infty} \cos\left(\frac{\pi}{2^{n+1}}\right) = \text{limit}_{\theta \to 0}\, \cos \theta = \cos 0 = 1\)


Equating the two, and multiplying both sides by a factor of \(\displaystyle 2\) then gives the following infinitely nested radical:



\(\displaystyle \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2\)

Comments and/or questions should be posted here:

http://mathhelpboards.com/commentar...-cosines-infinitely-nested-radicals-8283.html
 
Last edited by a moderator:

DreamWeaver

Well-known member
Sep 16, 2013
337
From the Cosine values listed above, we can deduce the equivalent forms for \(\displaystyle \sin(\pi/2^{n+1})\) in a number of different ways, for example, by writing \(\displaystyle \cos^2 x+\sin^2 x=1\) in the form:


\(\displaystyle \sin x=+\sqrt{1-\cos^2 x}\)


The square root is obviously positive, since we are working in the interval \(\displaystyle 0 < x < \pi/2\), where both the sine and cosine are strictly positive.


\(\displaystyle \sin\left(\frac{\pi}{2}\right) = 1\)


\(\displaystyle \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)


\(\displaystyle \sin\left(\frac{\pi}{8}\right) = \frac{1}{2}\, \sqrt{ 2-\sqrt{2} }\)


\(\displaystyle \sin\left(\frac{\pi}{16}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{2} } }\)


\(\displaystyle \sin\left(\frac{\pi}{32}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{2} } } }\)


\(\displaystyle \sin\left(\frac{\pi}{64}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{2} } } } }\)


\(\displaystyle \sin\left(\frac{\pi}{128}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{2} } } } } }\)



Using the simple formula \(\displaystyle \tan x= \sin x / \cos x\) we also have the following expressions for the Tangent function:



\(\displaystyle \tan\left(\frac{\pi}{4}\right)= 1\)


\(\displaystyle \tan\left(\frac{\pi}{8}\right)= \frac{
\sqrt{ 2 - \sqrt{2} }
}{
\sqrt{ 2 + \sqrt{2} }
}\)


\(\displaystyle \tan\left(\frac{\pi}{16}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{2} } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }
}\)


\(\displaystyle \tan\left(\frac{\pi}{32}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }
}\)


\(\displaystyle \tan\left(\frac{\pi}{64}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }
}\)


\(\displaystyle \tan\left(\frac{\pi}{128}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }
}\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
To develop further values, we could use the Addition Formulae for the Sine and Cosine:


\(\displaystyle \cos(x \pm y) = \cos x\cos y \mp \sin x\sin y\)


\(\displaystyle \sin(x \pm y) = \sin x\cos y \pm \cos x\sin y\)


For example, we could add consecutive pairs of values to obtain:


\(\displaystyle \cos\left( \frac{3\pi}{2^{n+1}} \right) =\cos\left( \frac{\pi}{2^n} + \frac{\pi}{2^{n+1}} \right)\)


\(\displaystyle \sin\left( \frac{3\pi}{2^{n+1}} \right) =\sin\left( \frac{\pi}{2^n} + \frac{\pi}{2^{n+1}} \right)\)


However, the's a more elegant way, which itself uses the addition formulae above; the following identity is easily proven by expanding the RHS:


\(\displaystyle \cos^2x-\cos^2y=\sin(x+y)\sin(y-x)\)


\(\displaystyle \Rightarrow\)


\(\displaystyle \cos^2\left( \frac{\pi}{8} \right)-\cos^2\left( \frac{3\pi}{8} \right) = \sin\left( \frac{\pi}{2} \right)\, \sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \)


Hence


\(\displaystyle \cos^2\left( \frac{3\pi}{8} \right) = \cos^2\left( \frac{\pi}{8} \right)- \frac{1}{\sqrt{2}}\)


and


\(\displaystyle \cos \left( \frac{3\pi}{8} \right) = \sqrt{ \frac{2+ \sqrt{2} }{4} - \frac{1}{\sqrt{2}} } = \)


\(\displaystyle \sqrt{ \frac{2+ \sqrt{2} }{4} - \frac{2\sqrt{2} }{ 4 } } = \sqrt{ \frac{2-\sqrt{2} }{4} } = \frac{1}{2} \sqrt{ 2-\sqrt{2} }\)


Finally, we use (04) repeatedly on \(\displaystyle \cos(3\pi/8)\) to obtain:



\(\displaystyle \cos \left( \frac{3\pi}{8} \right) = \frac{1}{2}\, \sqrt{ 2 -\sqrt{2} }\)


\(\displaystyle \cos \left( \frac{3\pi}{16} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } }\)


\(\displaystyle \cos \left( \frac{3\pi}{32} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } } }\)


\(\displaystyle \cos \left( \frac{3\pi}{64} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } } } }\)


\(\displaystyle \cos \left( \frac{3\pi}{128} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } } } } }\)



Finally, note that although \(\displaystyle \cos(3\pi/2^{n+1})\) approaches the limiting value \(\displaystyle 1\) slightly more slowly than \(\displaystyle \cos(\pi/2^{n+1})\) does, as \(\displaystyle n\) tends to infinity, the limit will ultimately be the same. Hence, with minimal effort, we conclude the following infinitely-nested radical evaluation:


\(\displaystyle \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \cdots \, + \sqrt{ 2-\sqrt{2} } } } } } = 2\)




More fun 'n' games shortly... (Heidy)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
We'll move onto some more interesting radicals shortly, but for now, a little house-keeping - as it were...


By the addition formula for the Cosine, namely \(\displaystyle \cos(x \pm y) =\cos x\cos y \mp \sin x\sin y\), we have:


\(\displaystyle \cos(\pi-\theta) = -\cos \theta\)


Let \(\displaystyle \theta = (\pi/2^{n+1})\), then


\(\displaystyle \cos\left( \pi- \frac{\pi}{2^{n+1}} \right) = \cos \left( \frac{(2^{n+1}-1)\pi}{2^{n+1}} \right) = - \cos \left( \frac{\pi}{2^{n+1}} \right)\)


Applying this identity to the Cosine values in the first post of this thread gives:


\(\displaystyle \cos\left( \frac{3\, \pi}{4} \right) = -\frac{ \sqrt{2} }{2}\)


\(\displaystyle \cos\left( \frac{7\, \pi}{8} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{2} }\)


\(\displaystyle \cos\left( \frac{15\, \pi}{16} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }\)


\(\displaystyle \cos\left( \frac{31\, \pi}{32} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }\)


\(\displaystyle \cos\left( \frac{63\, \pi}{64} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }\)


\(\displaystyle \cos\left( \frac{127\, \pi}{128} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }\)




A similar approach to the Cosine values in the previous post gives:



\(\displaystyle \cos\left( \frac{5\, \pi}{8} \right) = - \frac{1}{2}\, \sqrt{ 2 - \sqrt{2} }\)


\(\displaystyle \cos\left( \frac{13\, \pi}{16} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } }\)


\(\displaystyle \cos\left( \frac{29\, \pi}{32} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } }\)


\(\displaystyle \cos\left( \frac{61\, \pi}{64} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } } }\)


\(\displaystyle \cos\left( \frac{125\, \pi}{128} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } } } }\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
To obtain evaluations for \(\displaystyle \sin(\pi/3(2^{n+1}))\) and \(\displaystyle \cos(\pi/3(2^{n+1}))\) - from the values we already have - we convert the duplication formula for the Sine into a triplication formula. Explicitly, by writing \(\displaystyle \sin 3x = \sin(x+2x)\) and then expanding the RHS, it is possible to show that:


\(\displaystyle \sin 3x = 3\sin x - 4\sin^3x\)


Setting \(\displaystyle x=\pi/3\) then gives


\(\displaystyle \sin \pi = 0 = \sin\left(\frac{\pi}{3}\right)\, \Bigg[3-4\sin^2\left(\frac{\pi}{3}\right)
\Bigg]\)


By considering a right-angled triangle, clearly \(\displaystyle sin(\pi/3) \ne 0\), so we need to solve the quadratic (in the big square brackets):


\(\displaystyle 3-4\sin^2\left(\frac{\pi}{3}\right) =0 \Leftrightarrow \sin\left(\frac{\pi}{3}\right) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}\)


As before, we are working in the interval \(\displaystyle 0 < x < \pi\), where both the Sine and Cosine are strictly positive, so we took the positive square root of \(\displaystyle 3/4\).

The elementary identity \(\displaystyle \cos^2x+\sin^2x = 1\) then gives:


\(\displaystyle \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)


Finally, we use that last value in the RHS of (04) to obtain the series of values:


\(\displaystyle \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)


\(\displaystyle \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)


\(\displaystyle \cos\left(\frac{\pi}{12}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 3 } }\)


\(\displaystyle \cos\left(\frac{\pi}{24}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }\)


\(\displaystyle \cos\left(\frac{\pi}{48}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }\)


\(\displaystyle \cos\left(\frac{\pi}{96}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }\)


\(\displaystyle \cos\left(\frac{\pi}{192}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }\)


\(\displaystyle \cos\left(\frac{\pi}{384}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }\)



And since \(\displaystyle \cos(\pi-\theta) \equiv -\cos \theta\), we also have






\(\displaystyle \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\)


\(\displaystyle \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\)


\(\displaystyle \cos\left(\frac{11\pi}{12}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 3 } }\)


\(\displaystyle \cos\left(\frac{23\pi}{24}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }\)


\(\displaystyle \cos\left(\frac{47\pi}{48}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }\)


\(\displaystyle \cos\left(\frac{95\pi}{96}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }\)


\(\displaystyle \cos\left(\frac{191\pi}{192}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }\)


\(\displaystyle \cos\left(\frac{383\pi}{384}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }\)


From the addition formula for the Sine, \(\displaystyle \sin(\pi-\theta) = \sin \theta\). Hence, from the first group of Cosine values above, and the identity \(\displaystyle cos^2x+\sin^2x=1\), we have:


\(\displaystyle \sin\left(\frac{\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\)


\(\displaystyle \sin\left(\frac{\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)


\(\displaystyle \sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{11\pi}{12}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 3 } }\)


\(\displaystyle \sin\left(\frac{\pi}{24}\right) = \sin\left(\frac{23\pi}{24}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }\)


\(\displaystyle \sin\left(\frac{\pi}{48}\right) = \sin\left(\frac{47\pi}{48}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }\)


\(\displaystyle \sin\left(\frac{\pi}{96}\right) = \sin\left(\frac{95\pi}{96}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }\)


\(\displaystyle \sin\left(\frac{\pi}{192}\right) = \sin\left(\frac{191\pi}{192}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }\)


\(\displaystyle \sin\left(\frac{\pi}{384}\right) = \sin\left(\frac{383\pi}{384}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
From the previous values, and the definition of the Tangent as \(\displaystyle \tan x= \sin x/\cos x\) we easily obtain:



\(\displaystyle \tan\left(\frac{\pi}{3}\right) = \sqrt{3}\)


\(\displaystyle \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\)


\(\displaystyle \tan\left(\frac{\pi}{12}\right) = \frac{
\sqrt{ 2- \sqrt{ 3 } }
}{
\sqrt{ 2+ \sqrt{ 3 } }
}\)


\(\displaystyle \tan\left(\frac{\pi}{24}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }
}\)


\(\displaystyle \tan\left(\frac{\pi}{48}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}\)


\(\displaystyle \tan\left(\frac{\pi}{96}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}\)


\(\displaystyle \tan\left(\frac{\pi}{192}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}
\)


\(\displaystyle \tan\left(\frac{\pi}{384}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}
\)







\(\displaystyle \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}\)


\(\displaystyle \tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}\)


\(\displaystyle \tan\left(\frac{11\pi}{12}\right) = -\frac{
\sqrt{ 2- \sqrt{ 3 } }
}{
\sqrt{ 2+ \sqrt{ 3 } }
}\)


\(\displaystyle \tan\left(\frac{23\pi}{24}\right) = -\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }
}\)


\(\displaystyle \tan\left(\frac{47\pi}{48}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}\)


\(\displaystyle \tan\left(\frac{95\pi}{96}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}\)


\(\displaystyle \tan\left(\frac{191\pi}{192}\right) = -\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}
\)


\(\displaystyle \tan\left(\frac{383\pi}{384}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}
\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
In light of all of the previous examples, it's clear that we should be able to represent the Sine, Cosine, and Tangent at \(\displaystyle \theta = \pi/(2^{m+1}q)\) in a generalized - multiply-nested - closed form. This is indeed the case.

We define the "Fundamental Argument":

\(\displaystyle \varphi = \frac{\pi}{q}\)


Where \(\displaystyle q\) is a prime number:


\(\displaystyle q \in \{1, 2, 3, 5, 7, 11, 13, \cdots\, \}\)


By formula (04) we have:


\(\displaystyle \cos\left(\frac{\varphi}{2}\right) = \sqrt{ \frac{1+\cos \varphi}{2} }=\sqrt{ \frac{2}{4}+\frac{2\cos \varphi}{4} }= \frac{1}{2} \sqrt{ 2+2\cos \varphi }\)


Similarly,


\(\displaystyle \cos\left(\frac{\varphi}{4}\right) = \frac{1}{2} \sqrt{ 2+2\cos (\varphi/2) }=\)


\(\displaystyle \frac{1}{2} \sqrt{ 2+2\left( \frac{1}{2} \sqrt{ 2+2\cos \varphi } \right) }=\frac{1}{2} \sqrt{ 2+ \sqrt{ 2+2\cos \varphi } }\)


and more generally, iteration leads to


\(\displaystyle \cos\left(\frac{\varphi}{2^n}\right) = \frac{1}{2} \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}\)


Where the \(\displaystyle (n)\) superscript on the RHS represents the 'degree of nesting', ie the number of successive square roots. An equivalent form for the Sine is found by applying \(\displaystyle \sin^2x=1-\cos^2x\) to the previous expression:


\(\displaystyle \sin\left(\frac{\varphi}{2^n}\right) = \sqrt{ 1- \Bigg[ \frac{1}{2} \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)} \, \Bigg]^2 }= \)


\(\displaystyle \sqrt{ \frac{4}{4} - \frac{1}{4} \Bigg[ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } \, \Bigg|^{(n-1)} \, \Bigg] }= \)


\(\displaystyle \frac{1}{2} \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}\)



Combining the two, we obtain the following expression for the Tangent:


\(\displaystyle \tan\left(\frac{\varphi}{2^n}\right) =
\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}
}\)
 
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