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

2 Answers

Answer:

#f(n) = varphi^n(cos pi)#
#g(n) = psi(varphi^n(cos 2 pi))#

Explanation:

#0 <=x <= pi Rightarrow cos frac{x}{2} = sqrt {(1 + cos x)/2} = varphi(cos x)#

#0 <=x <= 2 pi Rightarrow sin frac{x}{2} = sqrt {(1 - cos x)/2} = psi(cos x)#

#varphi(t) = sqrt {(1 + t)/2}#
#psi(t) = sqrt {(1 - t)/2}#

#f(0) = cos pi#
#f(1) = varphi(cos pi)#
#f(2) = varphi(cos frac{pi}{2}) = varphi(varphi(cos pi)) = varphi^2(cos pi)#

#f(n) = varphi^n(cos pi)#

#g(0) = sin pi = psi(cos 2pi)#
#g(1) = sin frac{pi}{2} = psi(cos pi) = psi(varphi(cos 2 pi))#
#g(2) = psi(cos frac{pi}{2}) = psi(varphi(cos pi)) = psi(varphi^2(cos 2 pi))#

#g(n) = psi(varphi^n(cos 2 pi))#

More simple than this, do you have?

Answer:

See below.

Explanation:

Using the identities

#cos(a+b)=cosa cosb-sina sinb#
#sin(a+b)=sina cosb+cosa sinb#

we have

#cos(pi/2^(n-1)) = cos^2(pi/2^n)-sin^2(pi/2^n)#
#sin(pi/2^(n-1)) = 2cos(pi/2^n)sin(pi/2^n)#

Solving for #sin(pi/2^n), cos(pi/2^n)# we have

#sin(pi/2^n) = sqrt((1-cos(pi/2^(n-1)))/2)#
#cos(pi/2^n) = sin(pi/2^(n-1))/sqrt(2(1-cos(pi/2^(n-1))))#

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))#