Theorem: Dilogarithm of One Minus Reciprocal of Golden Mean

:$\map {\Li_2} {1 - \dfrac 1 \phi} = \dfrac 2 5 \map \zeta 2 - \paren {\map \ln \phi}^2$

Proof:
We first note the following:
{{begin-eqn}}
{{eqn | n = 1
      | l = -\frac 1 \phi
      | r = 1 - \phi
      | c = Reciprocal Form of One Minus Golden Mean
}}
{{eqn | ll= \leadsto
      | n = 2
      | l = \frac 1 {\phi^2}
      | r = 1 - \dfrac 1 \phi
      | c = dividing through by $-\phi$ and rearranging
}}
{{end-eqn}}
We now note:
{{begin-eqn}}
{{eqn | l = \map {\Li_2} z + \map {\Li_2} {-z}
      | r = \dfrac 1 2 \map {\Li_2} {z^2}
      | c = Dilogarithm of Square
}}
{{eqn | ll= \leadsto
      | l = \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {-\frac 1 \phi}
      | r = \dfrac 1 2 \map {\Li_2} {\frac 1 {\phi^2} }
      | c = setting $z := \dfrac 1 \phi$
}}
{{eqn | ll= \leadsto
      | n = 3
      | l = \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {1 - \phi} - \dfrac 1 2 \map {\Li_2} {1 - \dfrac 1 \phi}
      | r = 0
      | c = substituting from $(1)$ and $(2)$ above
}}
{{eqn | l = \map {\Li_2} {1 - z} + \map {\Li_2} {1 - \dfrac 1 z}
      | r = -\dfrac 1 2 \map {\ln^2} z
      | c = Dilogarithm of One Minus Z Plus Dilogarithm of One Minus Reciprocal of Z
}}
{{eqn | ll= \leadsto
      | l = \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {1 - \dfrac 1 {\frac 1 \phi} }
      | r = -\dfrac 1 2 \map {\ln^2} {\frac 1 \phi}
      | c = setting $z := \dfrac 1 \phi$
}}
{{eqn | ll= \leadsto
      | l = \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {1 - \phi}
      | r = -\dfrac 1 2 \paren {\map \ln 1 - \map \ln \phi}^2
      | c = Difference of Logarithms
}}
{{eqn | ll= \leadsto
      | n = 4
      | l = \map {\Li_2} {1 - \frac 1 \phi} + \map {\Li_2} {1 - \phi}
      | r = -\dfrac 1 2 \map {\ln^2} \phi
      | c = Natural Logarithm of 1 is 0
}}
{{eqn | ll= \leadsto
      | n = 5
      | l = \map {\Li_2} {\frac 1 \phi} - \dfrac 3 2 \map {\Li_2} {1 - \dfrac 1 \phi}
      | r = \dfrac 1 2 \map {\ln^2} \phi
      | c = $(3) - (4)$
}}
{{eqn | l = \map {\Li_2} z + \map {\Li_2} {1 - z}
      | r = \map \zeta 2 - \map \ln z \map \ln {1 - z}
      | c = Dilogarithm Reflection Formula
}}
{{eqn | ll= \leadsto
      | l = \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {1 - \frac 1 \phi}
      | r = \map \zeta 2 - \map \ln {\frac 1 \phi} \map \ln {1 - \frac 1 \phi}
      | c = $z := \dfrac 1 \phi$
}}
{{eqn | r = \map \zeta 2 - \map \ln {\frac 1 \phi} \map \ln {\frac 1 {\phi^2} }
      | c = 
}}
{{eqn | r = \map \zeta 2 - \paren {\map \ln 1 - \map \ln \phi} \paren {\map \ln 1 - \map \ln {\phi^2} }
      | c = Difference of Logarithms
}}
{{eqn | r = \map \zeta 2 + \map \ln \phi \paren {-2 \map \ln \phi}
      | c = Natural Logarithm of 1 is 0
}}
{{eqn | r = \map \zeta 2 - 2 \paren {\map \ln \phi}^2
      | c = 
}}
{{eqn | ll = \leadsto
      | n = 6
      | l = \map {\Li_2} {\frac 1 \phi} + \map {\Li_2} {1 - \frac 1 \phi}
      | r = \map \zeta 2 - 2 \paren {\map \ln \phi}^2
      | c = 
}}
{{eqn | ll = \leadsto
      | l = -\dfrac 5 2 \map {\Li_2} {1 - \dfrac 1 \phi}
      | r = \dfrac 1 2 \map {\ln^2} \phi - \paren {\map \zeta 2 - 2 \paren {\map \ln \phi}^2}
      | c = $(5) - (6)$
}}
{{eqn | ll = \leadsto
      | l = \map {\Li_2} {1 - \dfrac 1 \phi}
      | r = \dfrac 2 5 \map \zeta 2 - \paren {\map \ln \phi}^2
      | c = multiplying through by $-\dfrac 2 5$ and rearranging
}}
{{end-eqn}}
{{qed}}