# Have we formulas for f(n) = cos frac{pi}{2^n} and g(n) = sin frac{pi}{2^n} in radicals?

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#### Explanation

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#### Explanation:

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1
May 31, 2017

See below.

#### Explanation:

Using the identities

$\cos \left(a + b\right) = \cos a \cos b - \sin a \sin b$
$\sin \left(a + b\right) = \sin a \cos b + \cos a \sin b$

we have

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

Solving for $\sin \left(\frac{\pi}{2} ^ n\right) , \cos \left(\frac{\pi}{2} ^ n\right)$ we have

$\sin \left(\frac{\pi}{2} ^ n\right) = \sqrt{\frac{1 - \cos \left(\frac{\pi}{2} ^ \left(n - 1\right)\right)}{2}}$
$\cos \left(\frac{\pi}{2} ^ n\right) = \sin \frac{\frac{\pi}{2} ^ \left(n - 1\right)}{\sqrt{2 \left(1 - \cos \left(\frac{\pi}{2} ^ \left(n - 1\right)\right)\right)}}$

and computing some samples...

{(n, cos(pi/2^n), sin(pi/2^n)), (0, -1, 0), (1, 1/2 sqrt[2], 1/2 sqrt[2]), (2, 1/2sqrt[2 + sqrt[2]], 1/2sqrt[2 - sqrt[2]]), (3, 1/2 sqrt[2 + sqrt[2 + sqrt[2]]], 1/2 sqrt[2 - sqrt[2 + sqrt[2]]]), (4, 1/2 sqrt[2 + sqrt[2 + sqrt[2 + sqrt[2]]]], 1/2 sqrt[2 - sqrt[2 + sqrt[2 + sqrt[2]]]]),(cdots, cdots,cdots))

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

0
May 30, 2017

$f \left(n\right) = {\varphi}^{n} \left(\cos \pi\right)$
$g \left(n\right) = \psi \left({\varphi}^{n} \left(\cos 2 \pi\right)\right)$

#### Explanation:

$0 \le x \le \pi R i g h t a r r o w \cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \varphi \left(\cos x\right)$

$0 \le x \le 2 \pi R i g h t a r r o w \sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \psi \left(\cos x\right)$

$\varphi \left(t\right) = \sqrt{\frac{1 + t}{2}}$
$\psi \left(t\right) = \sqrt{\frac{1 - t}{2}}$

$f \left(0\right) = \cos \pi$
$f \left(1\right) = \varphi \left(\cos \pi\right)$
$f \left(2\right) = \varphi \left(\cos \frac{\pi}{2}\right) = \varphi \left(\varphi \left(\cos \pi\right)\right) = {\varphi}^{2} \left(\cos \pi\right)$

$f \left(n\right) = {\varphi}^{n} \left(\cos \pi\right)$

$g \left(0\right) = \sin \pi = \psi \left(\cos 2 \pi\right)$
$g \left(1\right) = \sin \frac{\pi}{2} = \psi \left(\cos \pi\right) = \psi \left(\varphi \left(\cos 2 \pi\right)\right)$
$g \left(2\right) = \psi \left(\cos \frac{\pi}{2}\right) = \psi \left(\varphi \left(\cos \pi\right)\right) = \psi \left({\varphi}^{2} \left(\cos 2 \pi\right)\right)$

$g \left(n\right) = \psi \left({\varphi}^{n} \left(\cos 2 \pi\right)\right)$

More simple than this, do you have?

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