Closed form of $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\varphi-1)$ I am trying to calculate the dilogarithm of the golden ratio and its conjugate $\Phi = \varphi-1$. Eg the solutions of the equation $u^2 - u = 1$. 
From Wikipdia one has the following

\begin{align*} 
  \operatorname{Li}_2\left( \frac{1 + \sqrt{5}}{2} \right)
   & = -\int_0^\varphi \frac{\log(1-t)}{t}\,\mathrm{d}t 
     =  \phantom{-}\frac{\pi^2}{10} - \log^2\left( \Phi\right) \\
  \operatorname{Li}_2\left( \frac{1 - \sqrt{5}}{2} \right)
   & = -\int_0^\Phi \frac{\log(1-t)}{t}\,\mathrm{d}t 
     =  -\frac{\pi^2}{15} - \log^2\left( -\Phi\right) \\
 \end{align*}

I am quite certain that these two special values can be shown by combining the identites for the dilogarithm, and forming a system of equations. But I am having some problems obtaining a set of equations only involving $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\Phi)$. Can anyone show me how to set up the system of equations from the identites, or perhaps a different path in showing these two values? 

This solution is very similar to @Start wearing purple's solution above.
Using the following identities (check Eq $3$,$4$ and $5$).
$$\text{Li}_2(1-x)+\text{Li}_2\left(\frac{x-1}{x}\right)=-\frac12\ln^2\left(\frac{1}{x}\right)\tag1$$
$$\text{Li}_2(x)+\text{Li}_2(-x)-\frac12\text{Li}_2(x^2)=0\tag2$$
$$\text{Li}_2(x)+\text{Li}_2(1-x)=\zeta(2)-\ln(x)\ln(1-x)\tag3$$
If we set $x=\frac{\sqrt{5}-1}{2}$ we notice that 
$$1-x=\frac{3-\sqrt{5}}{2}$$
$$\frac{x-1}{x}=\frac{1-\sqrt{5}}{2}$$
$$\frac1x=\frac{1+\sqrt{5}}{2}$$
$$x^2=\frac{3-\sqrt{5}}{2}$$
$$\ln(x)\ln(1-x)=2\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
Plugging these values in $(1)$, $(2)$ and $(3)$ we get
$$\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}+\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}=-\frac12\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)+\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}-\frac12\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=0$$
$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)+\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=\zeta(2)-2\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
Solving this system of equations, we have 
$$\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=\frac{\pi^2}{15}-\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
$$\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}=-\frac{\pi^2}{15}+\frac12\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)=\frac{\pi^2}{10}-\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$
