Sines, Cosines, and infinitely nested radicals

[DreamWeaver]

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 $$\cos(\pi/2^kn),\, \sin(\pi/2^kn), \,$$ and $$\tan(\pi/2^kn) $$ in closed form, for ever smaller arguments $$(k, n \in \mathbb{Z}^{+})$$.

(2) By considering the limiting values of $$\cos(\pi/2^kn),\, \sin(\pi/2^kn), \,$$ and $$\tan(\pi/2^kn) $$ as $$k\to \infty$$, we evaluate a number of infinitely-nested radicals, such as:$$\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2$$(3) Finally, we consider three new (?) trigonometric functions, which I will henceforth refer to as the Fractional Sine, Cosine, and Tangent functions respectively:$$\mathscr{Fs}(\theta) = \sum_{k=0}^{\infty} \sin\left( \frac{\theta}{2^k} \right)$$$$\mathscr{Fc}(\theta) = \sum_{k=0}^{\infty} (-1)^k\cos\left( \frac{\theta}{2^k} \right)$$$$\mathscr{Ft}(\theta) = \sum_{k=0}^{\infty} \tan\left( \frac{\theta}{2^k} \right)$$Naturally, some restrictions will need to be applied to $$\theta$$ in each case, to ensure convergence. Also, note that the Fractional Cosine function is an alternating series; this is due to the fact that $$\lim_{\theta \to 0}\, \cos (\theta) = 1$$.
--------------------------------------------------

For now, we will assume the following two values of the Cosine:$$(01) \quad \cos\left(\frac{\pi}{2}\right) = 0$$$$(02) \quad \cos\left(\frac{\pi}{5}\right) = \frac{1}{2} \sqrt{ \frac{3+\sqrt{5} }{2} }$$In addition, the trigonometric identities used herein are elementary, so proofs will be omitted. The main identities used, aside from $$\cos^2 x+ \sin^2 x = 1$$, are essentially three different forms of the Cosine Double Angle formula:$$(03)\quad \sin\frac{x}{2} = \pm \sqrt{\frac{1-\cos x}{2}}$$$$(04)\quad \cos\frac{x}{2} = \pm \sqrt{\frac{1+\cos x}{2}}$$$$(05)\quad \tan^2x= \frac{1-\cos 2x}{1+\cos 2x}$$
In identity (03), we have $$\text{sgn}= + \Leftrightarrow 0 < x < 2\pi$$, whereas in identity (04) we have $$\text{sgn}= + \Leftrightarrow -\pi < x < \pi$$.
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Entering $$\cos(\pi/2)= 0$$ in the RHS of identity (04) gives:$$\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$$If we repeat this process, each time entering our new value in the RHS of identity (04), we obtain the series of values:$$\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$$$\cos\left(\frac{\pi}{8}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{2} }$$$$\cos\left(\frac{\pi}{16}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }$$$$\cos\left(\frac{\pi}{32}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }$$$$\cos\left(\frac{\pi}{64}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } }$$$$\cos\left(\frac{\pi}{128}\right) = \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 } } } } } }$$
We introduce the operator $$\Delta_n(x)$$, where the operation is "add 2 and then square-root the sum" $$n$$-times. Here, $$x$$ is the operand, that is the initial value that is to be operated on. Then we man write:$$\cos\left(\frac{\pi}{2^{n+1}}\right) = \frac{1}{2}\, \Delta_n(0)\quad \quad \quad n=0, 1, 2, \cdots $$Now, on the one hand, we have$$\text{limit}_{n\to \infty} \, \Delta_n(0) = \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } }$$whereas on the other, $$\text{limit}_{n\to \infty} \cos\left(\frac{\pi}{2^{n+1}}\right) = \text{limit}_{\theta \to 0}\, \cos \theta = \cos 0 = 1$$Equating the two, and multiplying both sides by a factor of $$2$$ then gives the following infinitely nested radical:
$$\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \cdots \, } } } } } } = 2$$

Comments and/or questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-sines-cosines-infinitely-nested-radicals-8283.html

 

[DreamWeaver]

From the Cosine values listed above, we can deduce the equivalent forms for $$\sin(\pi/2^{n+1})$$ in a number of different ways, for example, by writing $$\cos^2 x+\sin^2 x=1$$ in the form:$$\sin x=+\sqrt{1-\cos^2 x}$$The square root is obviously positive, since we are working in the interval $$0 < x < \pi/2$$, where both the sine and cosine are strictly positive.$$\sin\left(\frac{\pi}{2}\right) = 1$$$$\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$$$\sin\left(\frac{\pi}{8}\right) = \frac{1}{2}\, \sqrt{ 2-\sqrt{2} }$$$$\sin\left(\frac{\pi}{16}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{2} } }$$$$\sin\left(\frac{\pi}{32}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{2} } } }$$$$\sin\left(\frac{\pi}{64}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{2} } } } }$$$$\sin\left(\frac{\pi}{128}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{2} } } } } }$$
Using the simple formula $$\tan x= \sin x / \cos x$$ we also have the following expressions for the Tangent function:
$$\tan\left(\frac{\pi}{4}\right)= 1$$$$\tan\left(\frac{\pi}{8}\right)= \frac{
\sqrt{ 2 - \sqrt{2} }
}{
\sqrt{ 2 + \sqrt{2} }
}$$$$\tan\left(\frac{\pi}{16}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{2} } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }
}$$$$\tan\left(\frac{\pi}{32}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }
}$$$$\tan\left(\frac{\pi}{64}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }
}$$$$\tan\left(\frac{\pi}{128}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }
}$$

 

[DreamWeaver]

To develop further values, we could use the Addition Formulae for the Sine and Cosine:$$\cos(x \pm y) = \cos x\cos y \mp \sin x\sin y$$$$\sin(x \pm y) = \sin x\cos y \pm \cos x\sin y$$For example, we could add consecutive pairs of values to obtain:$$\cos\left( \frac{3\pi}{2^{n+1}} \right) =\cos\left( \frac{\pi}{2^n} + \frac{\pi}{2^{n+1}} \right)$$$$\sin\left( \frac{3\pi}{2^{n+1}} \right) =\sin\left( \frac{\pi}{2^n} + \frac{\pi}{2^{n+1}} \right)$$However, the's a more elegant way, which itself uses the addition formulae above; the following identity is easily proven by expanding the RHS:$$\cos^2x-\cos^2y=\sin(x+y)\sin(y-x)$$$$\Rightarrow$$$$\cos^2\left( \frac{\pi}{8} \right)-\cos^2\left( \frac{3\pi}{8} \right) = \sin\left( \frac{\pi}{2} \right)\, \sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} $$Hence$$\cos^2\left( \frac{3\pi}{8} \right) = \cos^2\left( \frac{\pi}{8} \right)- \frac{1}{\sqrt{2}}$$and $$\cos \left( \frac{3\pi}{8} \right) = \sqrt{ \frac{2+ \sqrt{2} }{4} - \frac{1}{\sqrt{2}} } = $$$$\sqrt{ \frac{2+ \sqrt{2} }{4} - \frac{2\sqrt{2} }{ 4 } } = \sqrt{ \frac{2-\sqrt{2} }{4} } = \frac{1}{2} \sqrt{ 2-\sqrt{2} }$$Finally, we use (04) repeatedly on $$\cos(3\pi/8)$$ to obtain:
$$\cos \left( \frac{3\pi}{8} \right) = \frac{1}{2}\, \sqrt{ 2 -\sqrt{2} }$$$$\cos \left( \frac{3\pi}{16} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } }$$$$\cos \left( \frac{3\pi}{32} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } } }$$$$\cos \left( \frac{3\pi}{64} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } } } }$$$$\cos \left( \frac{3\pi}{128} \right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2-\sqrt{2} } } } } }$$
Finally, note that although $$\cos(3\pi/2^{n+1})$$ approaches the limiting value $$1$$ slightly more slowly than $$\cos(\pi/2^{n+1})$$ does, as $$n$$ tends to infinity, the limit will ultimately be the same. Hence, with minimal effort, we conclude the following infinitely-nested radical evaluation:$$\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \cdots \, + \sqrt{ 2-\sqrt{2} } } } } } = 2$$

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

 

[DreamWeaver]

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 $$\cos(x \pm y) =\cos x\cos y \mp \sin x\sin y$$, we have:$$\cos(\pi-\theta) = -\cos \theta$$Let $$\theta = (\pi/2^{n+1})$$, then$$\cos\left( \pi- \frac{\pi}{2^{n+1}} \right) = \cos \left( \frac{(2^{n+1}-1)\pi}{2^{n+1}} \right) = - \cos \left( \frac{\pi}{2^{n+1}} \right)$$Applying this identity to the Cosine values in the first post of this thread gives:$$\cos\left( \frac{3\, \pi}{4} \right) = -\frac{ \sqrt{2} }{2}$$$$\cos\left( \frac{7\, \pi}{8} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{2} }$$$$\cos\left( \frac{15\, \pi}{16} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }$$$$\cos\left( \frac{31\, \pi}{32} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }$$$$\cos\left( \frac{63\, \pi}{64} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }$$$$\cos\left( \frac{127\, \pi}{128} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }$$

A similar approach to the Cosine values in the previous post gives:
$$\cos\left( \frac{5\, \pi}{8} \right) = - \frac{1}{2}\, \sqrt{ 2 - \sqrt{2} }$$$$\cos\left( \frac{13\, \pi}{16} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } }$$$$\cos\left( \frac{29\, \pi}{32} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } }$$$$\cos\left( \frac{61\, \pi}{64} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } } }$$$$\cos\left( \frac{125\, \pi}{128} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } } } }$$

 

[DreamWeaver]

To obtain evaluations for $$\sin(\pi/3(2^{n+1}))$$ and $$\cos(\pi/3(2^{n+1}))$$ - from the values we already have - we convert the duplication formula for the Sine into a triplication formula. Explicitly, by writing $$\sin 3x = \sin(x+2x)$$ and then expanding the RHS, it is possible to show that:$$\sin 3x = 3\sin x - 4\sin^3x$$Setting $$x=\pi/3$$ then gives$$\sin \pi = 0 = \sin\left(\frac{\pi}{3}\right)\, \Bigg[3-4\sin^2\left(\frac{\pi}{3}\right)
\Bigg]$$By considering a right-angled triangle, clearly $$sin(\pi/3) \ne 0$$, so we need to solve the quadratic (in the big square brackets):$$3-4\sin^2\left(\frac{\pi}{3}\right) =0 \Leftrightarrow \sin\left(\frac{\pi}{3}\right) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$As before, we are working in the interval $$0 < x < \pi$$, where both the Sine and Cosine are strictly positive, so we took the positive square root of $$3/4$$.

The elementary identity $$\cos^2x+\sin^2x = 1$$ then gives:$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$Finally, we use that last value in the RHS of (04) to obtain the series of values:$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$$$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$$$\cos\left(\frac{\pi}{12}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 3 } }$$$$\cos\left(\frac{\pi}{24}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }$$$$\cos\left(\frac{\pi}{48}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }$$$$\cos\left(\frac{\pi}{96}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }$$$$\cos\left(\frac{\pi}{192}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }$$$$\cos\left(\frac{\pi}{384}\right) = \frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }$$
And since $$\cos(\pi-\theta) \equiv -\cos \theta$$, we also have
$$\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$$$\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$$$\cos\left(\frac{11\pi}{12}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 3 } }$$$$\cos\left(\frac{23\pi}{24}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }$$$$\cos\left(\frac{47\pi}{48}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }$$$$\cos\left(\frac{95\pi}{96}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }$$$$\cos\left(\frac{191\pi}{192}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }$$$$\cos\left(\frac{383\pi}{384}\right) = -\frac{1}{2}\, \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }$$From the addition formula for the Sine, $$\sin(\pi-\theta) = \sin \theta$$. Hence, from the first group of Cosine values above, and the identity $$cos^2x+\sin^2x=1$$, we have:$$\sin\left(\frac{\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$$$$\sin\left(\frac{\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$$$$\sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{11\pi}{12}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 3 } }$$$$\sin\left(\frac{\pi}{24}\right) = \sin\left(\frac{23\pi}{24}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }$$$$\sin\left(\frac{\pi}{48}\right) = \sin\left(\frac{47\pi}{48}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }$$$$\sin\left(\frac{\pi}{96}\right) = \sin\left(\frac{95\pi}{96}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }$$$$\sin\left(\frac{\pi}{192}\right) = \sin\left(\frac{191\pi}{192}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }$$$$\sin\left(\frac{\pi}{384}\right) = \sin\left(\frac{383\pi}{384}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }$$

 

[DreamWeaver]

From the previous values, and the definition of the Tangent as $$\tan x= \sin x/\cos x$$ we easily obtain:
$$\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$$$$\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$$$$\tan\left(\frac{\pi}{12}\right) = \frac{
\sqrt{ 2- \sqrt{ 3 } }
}{
\sqrt{ 2+ \sqrt{ 3 } }
}$$$$\tan\left(\frac{\pi}{24}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }
}$$$$\tan\left(\frac{\pi}{48}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}$$$$\tan\left(\frac{\pi}{96}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}$$$$\tan\left(\frac{\pi}{192}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}
$$$$\tan\left(\frac{\pi}{384}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}
$$

$$\tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}$$$$\tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}$$$$\tan\left(\frac{11\pi}{12}\right) = -\frac{
\sqrt{ 2- \sqrt{ 3 } }
}{
\sqrt{ 2+ \sqrt{ 3 } }
}$$$$\tan\left(\frac{23\pi}{24}\right) = -\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }
}$$$$\tan\left(\frac{47\pi}{48}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}$$$$\tan\left(\frac{95\pi}{96}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}$$$$\tan\left(\frac{191\pi}{192}\right) = -\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}
$$$$\tan\left(\frac{383\pi}{384}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}
$$

 

[DreamWeaver]

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

We define the "Fundamental Argument":

$$\varphi = \frac{\pi}{q}$$Where $$q$$ is a prime number:$$q \in \{1, 2, 3, 5, 7, 11, 13, \cdots\, \}$$By formula (04) we have:$$\cos\left(\frac{\varphi}{2}\right) = \sqrt{ \frac{1+\cos \varphi}{2} }=\sqrt{ \frac{2}{4}+\frac{2\cos \varphi}{4} }= \frac{1}{2} \sqrt{ 2+2\cos \varphi }$$Similarly,$$\cos\left(\frac{\varphi}{4}\right) = \frac{1}{2} \sqrt{ 2+2\cos (\varphi/2) }=$$$$\frac{1}{2} \sqrt{ 2+2\left( \frac{1}{2} \sqrt{ 2+2\cos \varphi } \right) }=\frac{1}{2} \sqrt{ 2+ \sqrt{ 2+2\cos \varphi } }$$and more generally, iteration leads to$$\cos\left(\frac{\varphi}{2^n}\right) = \frac{1}{2} \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}$$Where the $$(n)$$ 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 $$\sin^2x=1-\cos^2x$$ to the previous expression:$$\sin\left(\frac{\varphi}{2^n}\right) = \sqrt{ 1- \Bigg[ \frac{1}{2} \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)} \, \Bigg]^2 }= $$ $$\sqrt{ \frac{4}{4} - \frac{1}{4} \Bigg[ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } \, \Bigg|^{(n-1)} \, \Bigg] }= $$$$\frac{1}{2} \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}$$
Combining the two, we obtain the following expression for the Tangent:$$\tan\left(\frac{\varphi}{2^n}\right) =
\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+\, \cdots \, + \sqrt{ 2+ 2\cos \varphi } } } } } \, \Bigg|^{(n)}
}$$