Sines, Cosines, and infinitely nested radicals

  • MHB
  • Thread starter DreamWeaver
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In summary: The first is the development of multiply-nested radicals for expressing the Cosine, Sine, and Tangent functions in closed form. The second topic focuses on evaluating infinitely-nested radicals by taking the limiting values of the trigonometric functions as k approaches infinity. Finally, the conversation introduces three new trigonometric functions known as the Fractional Sine, Cosine, and Tangent functions. These functions require certain restrictions to ensure convergence and the Fractional Cosine function is an alternating series. The conversation also includes examples of using these functions to find the values of Cosine, Sine, and Tangent for various angles.
  • #1
DreamWeaver
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0
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:

http://mathhelpboards.com/commentary-threads-53/commentary-sines-cosines-infinitely-nested-radicals-8283.html
 
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  • #2
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:
\(\displaystyle \tan\left(\frac{\pi}{4}\right)= 1\)\(\displaystyle \tan\left(\frac{\pi}{8}\right)= \frac{
\sqrt{ 2 - \sqrt{2} }
}{
\sqrt{ 2 + \sqrt{2} }
}\)\(\displaystyle \tan\left(\frac{\pi}{16}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{2} } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } }
}\)\(\displaystyle \tan\left(\frac{\pi}{32}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }
}\)\(\displaystyle \tan\left(\frac{\pi}{64}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } }
}\)\(\displaystyle \tan\left(\frac{\pi}{128}\right)= \frac{
\sqrt{ 2 - \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }
}{
\sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } } } }
}\)
 
  • #3
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)
 
  • #4
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:
\(\displaystyle \cos\left( \frac{5\, \pi}{8} \right) = - \frac{1}{2}\, \sqrt{ 2 - \sqrt{2} }\)\(\displaystyle \cos\left( \frac{13\, \pi}{16} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } }\)\(\displaystyle \cos\left( \frac{29\, \pi}{32} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } }\)\(\displaystyle \cos\left( \frac{61\, \pi}{64} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } } }\)\(\displaystyle \cos\left( \frac{125\, \pi}{128} \right) = - \frac{1}{2}\, \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 - \sqrt{2} } } } } }\)
 
  • #5
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:\(\displaystyle \sin\left(\frac{\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\)\(\displaystyle \sin\left(\frac{\pi}{6}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)\(\displaystyle \sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{11\pi}{12}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 3 } }\)\(\displaystyle \sin\left(\frac{\pi}{24}\right) = \sin\left(\frac{23\pi}{24}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }\)\(\displaystyle \sin\left(\frac{\pi}{48}\right) = \sin\left(\frac{47\pi}{48}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }\)\(\displaystyle \sin\left(\frac{\pi}{96}\right) = \sin\left(\frac{95\pi}{96}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }\)\(\displaystyle \sin\left(\frac{\pi}{192}\right) = \sin\left(\frac{191\pi}{192}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }\)\(\displaystyle \sin\left(\frac{\pi}{384}\right) = \sin\left(\frac{383\pi}{384}\right) = \frac{1}{2}\, \sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }\)
 
  • #6
From the previous values, and the definition of the Tangent as \(\displaystyle \tan x= \sin x/\cos x\) we easily obtain:
\(\displaystyle \tan\left(\frac{\pi}{3}\right) = \sqrt{3}\)\(\displaystyle \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\)\(\displaystyle \tan\left(\frac{\pi}{12}\right) = \frac{
\sqrt{ 2- \sqrt{ 3 } }
}{
\sqrt{ 2+ \sqrt{ 3 } }
}\)\(\displaystyle \tan\left(\frac{\pi}{24}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }
}\)\(\displaystyle \tan\left(\frac{\pi}{48}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}\)\(\displaystyle \tan\left(\frac{\pi}{96}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}\)\(\displaystyle \tan\left(\frac{\pi}{192}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}
\)\(\displaystyle \tan\left(\frac{\pi}{384}\right) = \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}
\)

\(\displaystyle \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}\)\(\displaystyle \tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}\)\(\displaystyle \tan\left(\frac{11\pi}{12}\right) = -\frac{
\sqrt{ 2- \sqrt{ 3 } }
}{
\sqrt{ 2+ \sqrt{ 3 } }
}\)\(\displaystyle \tan\left(\frac{23\pi}{24}\right) = -\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 3 } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } }
}\)\(\displaystyle \tan\left(\frac{47\pi}{48}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } }
}\)\(\displaystyle \tan\left(\frac{95\pi}{96}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } }
}\)\(\displaystyle \tan\left(\frac{191\pi}{192}\right) = -\frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } }
}
\)\(\displaystyle \tan\left(\frac{383\pi}{384}\right) =- \frac{
\sqrt{ 2- \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}{
\sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 2+ \sqrt{ 3 } } } } } } }
}
\)
 
  • #7
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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FAQ: Sines, Cosines, and infinitely nested radicals

1. What are sines and cosines?

Sines and cosines are two of the trigonometric functions used to relate the angles of a triangle to the lengths of its sides. Sines are ratios of the opposite side to the hypotenuse, while cosines are ratios of the adjacent side to the hypotenuse.

2. How are sines and cosines used in mathematics and science?

Sines and cosines are used in a wide range of fields, including physics, engineering, and astronomy. They are used to model and understand periodic phenomena, such as sound waves, electromagnetic waves, and planetary motion.

3. What are infinitely nested radicals?

Infinitely nested radicals are mathematical expressions that contain a nested or repeating pattern of radicals, where the number of radicals continues infinitely. They are often used to represent irrational numbers or to solve certain types of equations.

4. How do you simplify infinitely nested radicals?

To simplify infinitely nested radicals, you can use the properties of radicals, such as the product and quotient rules, to combine like terms and simplify the expression. You may also need to use techniques such as rationalizing the denominator to eliminate any square roots in the denominator.

5. What are some real-world applications of infinitely nested radicals?

Infinitely nested radicals have various applications in engineering, physics, and computer science. They are used to model and analyze complex systems, such as electrical circuits and fractal patterns. They are also used in encryption algorithms and in generating random numbers for simulations and games.

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