■正多角形の作図と原始根(その268)

[Q]cos2π/13+cos6π/13+cos18π/13=?

[Q]sin2π/13+sin6π/13+sin18π/13=?

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最初に3群に分けると

b2=a^2+a^3+a^11+a^10

b4=a^4+a^6+a^9+a^7

b1=a^8+a^12+a^5+a

b2b4=a^6+a^8+a^11+a^9+a^7+a^9+a^12+a^10+a^2+a^4+a^7+a^5+a+a^3+a^6+a^4

=2b4+b1+b2

b4b1=a^12+a^3+a^9+a^5+a+a^5+a^11+a^7+a^4+a^8+a+a^10+a^2+a^6+a^12+a^8

=2b1+b2+b4

b1b2=a^10+a+a^7+a^3+a^11+a^2+a^8+a^4+a^6+a^10+a^3+a^12+a^5+a^9+a^2+a^11

=2b2+b1+b4

b2b4+b4b1+b1b2=4(b1+b2+b4)=-4

b2b4b1=2b4b1+b1^2+b1b2=4b1+2b2+2b4+b1^2+2b2+b1+b4=5b1+4b2+3b4+b1^2

b1^2=a^3+a^11+a^10+a^2+2a^7+2+2a^9+2a^4+2+2a^6

b1^2=b2+2a^7+2+2a^9+2a^4+2+2a^6=b2+2b4+4

b2b4b1=5b1+4b2+3b4+b2+2b4+4=5(b1+b2+b4)+4=-1

b1,b2,b4はx^3+x^2-4x+1=0の3根

b1^2=b2+2b4+4=(b2+b4)+b4+4=-1-b1+b4+4=b4+3-b1

b1^2=b2+2b4+4=2(b2+b4)-b2+4=2(-1-b1)-b2+4=-b2+2-2b1

b1=(a+a^12)+(a^5+a^8)

c1=a+a^12,c2=a^5+a^8

c1c2=a^6+a^9+a^4+a^7=b4

c1,c2はx^2-b1x+b4=0の2根

x^2-b1x+b4=x^2-b1x+b1-3+b1^2=0

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x=(y-1/3)とおく

(y-1/3)^3+(y-1/3)^2-4(y-1/3)+1=0

y^3-y^2+1/3・y-1/27+y^2-2/3・y+1/9-4y+4/3+1=0

y^3+(1/3-2/3-4)y-1/27+1/9+7/3=0

y^3+(-13/3)y+(-1+3+63)/27=0

y^3+(-13/3)y+65/27=0

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 3次方程式:x^3=px+qの解は

  x=3√A+3√B

  A=q/2+√((q/2)^2−(p/3)^3)

  B=q/2−√((q/2)^2−(p/3)^3)

で与えられる.

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p=13/3,q=-65/27

A=-65/54+{(65/54)^2-(13/9)^3}^1/2

B=-65/54-{(65/54)^2-(13/9)^3}^1/2

A=-65/54+{13^2・(5)^2/4/9^3-13^2・13/9^3}^1/2

B=-65/54-{13^2・(5)^2/4/9^3-13^2・13/9^3}^1/2

A=-65/54+13/9・{(5)^2/4/9-13/9}^1/2

B=-65/54-13/9・{(5)^2/4/9-13/9}^1/2

A=-65/54+13/9・{(25-52)/36}^1/2

B=-65/54-13/9・{(25-52)/36}^1/2

A=-65/54+13/18・{-3}^1/2

B=-65/54-13/18・{-3}^1/2

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