# Sines, Cosines, and infinitely nested radicals

#### DreamWeaver

##### Well-known member
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 $$\displaystyle \lim_{\theta \to 0}\, \cos (\theta) = 1$$.

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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 $$\displaystyle \text{sgn}= + \Leftrightarrow -\pi < x < \pi$$.

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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:

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#### DreamWeaver

##### Well-known member
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
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... #### DreamWeaver

##### Well-known member
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
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
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
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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