Search Results for ""

Trigonometric functions of pi/p for p prime have an especially complicated Galois-minimal representation. In particular, the case cos(pi/23) requires approximately 500 MB of ...

cos(pi/(24)) = 1/2sqrt(2+sqrt(2+sqrt(3))) (1) cos((5pi)/(24)) = 1/2sqrt(2+sqrt(2-sqrt(3))) (2) cos((7pi)/(24)) = 1/2sqrt(2-sqrt(2-sqrt(3))) (3) cos((11pi)/(24)) = ...

Construction of the angle pi/3=60 degrees produces a 30-60-90 triangle, which has angles theta=pi/3 and theta/2=pi/6. From the above diagram, write y=sintheta for the ...

cos(pi/(30)) = 1/4sqrt(7+sqrt(5)+sqrt(6(5+sqrt(5)))) (1) cos((7pi)/(30)) = 1/4sqrt(7-sqrt(5)+sqrt(6(5-sqrt(5)))) (2) cos((11pi)/(30)) = 1/4sqrt(7+sqrt(5)-sqrt(6(5+sqrt(5)))) ...

cos(pi/(32)) = 1/2sqrt(2+sqrt(2+sqrt(2+sqrt(2)))) (1) cos((3pi)/(32)) = 1/2sqrt(2+sqrt(2+sqrt(2-sqrt(2)))) (2) cos((5pi)/(32)) = 1/2sqrt(2+sqrt(2-sqrt(2-sqrt(2)))) (3) ...

Construction of the angle pi/4=45 degrees produces an isosceles right triangle. Since the sides are equal, sin^2theta+cos^2theta=2sin^2theta=1, (1) so solving for ...

Construction of the angle pi/6=30 degrees produces a 30-60-90 triangle, which has angles theta=pi/6 and 2theta=pi/3. From the above diagram, write y=sintheta for the vertical ...

cos(pi/8) = 1/2sqrt(2+sqrt(2)) (1) cos((3pi)/8) = 1/2sqrt(2-sqrt(2)) (2) cot(pi/8) = 1+sqrt(2) (3) cot((3pi)/8) = sqrt(2)-1 (4) csc(pi/8) = sqrt(4+2sqrt(2)) (5) csc((3pi)/8) ...

A class of area-preserving maps of the form theta_(i+1) = theta_i+2pialpha(r_i) (1) r_(i+1) = r_i, (2) which maps circles into circles but with a twist resulting from the ...

f(x)=1-2x^2 for x in [-1,1]. Fixed points occur at x=-1, 1/2, and order 2 fixed points at x=(1+/-sqrt(5))/4. The natural invariant of the map is rho(y)=1/(pisqrt(1-y^2)).