Proof clarification $-\int_0^{\pi/12}\log(\tan t)dt=\frac23\mathrm G$ $\mathrm G$ is Catalan's constant.
I would like clarification on the following proof provided by @M.N.C.E.:
$$
\begin{align}
&\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\
=&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\
=&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\tag1\\
=&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\
&\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\tag2\\
=&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{\mathrm G}}\color{#E2062C}{-\frac{1}{2}\mathrm G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}\mathrm G}}\tag3\\
=&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)\mathrm G=\large{-\frac{2}{3}\mathrm G}
\end{align}$$

I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$. 


A log of my attempts to find an alternate proof for the identity in question.
I tried to represent $(1)$ as a hypergeometric series because the $1/(2n+1)^2$ terms are after all what give us the identity
$$_3F_2\left(\frac12,\frac12,1;\frac32,\frac32;-1\right)=\mathrm G$$
So I was hoping that $$\frac{\sin\left[\frac\pi6(2n+1)\right]}{\sin\left[\frac\pi6(2n+3)\right]}=-1$$
But alas this is not the case. I see now that I would be missing an extra factor of $2/3$.
Another, more realistic attempt of mine comes by noting that
$$\sin\left[\frac\pi6(2n+1)\right]=\frac{\sqrt3}2\sin\frac{\pi n}3+\frac12\cos\frac{\pi n}3$$
So, by letting $K$ be $-1\cdot$(the quantity $(1)$), we have that
$$K=\frac{\sqrt3}2S+\frac12C$$
Where $$S=\sum_{n\geq0}\frac{\sin\frac{\pi n}3}{(2n+1)^2}$$
and $$C=\sum_{n\geq0}\frac{\cos\frac{\pi n}3}{(2n+1)^2}$$
Although neither of these seem to have a 'nice' $_pF_q$ representation. I have run out of ideas for proving that $K=\frac23\mathrm G$.
 A: Firstly, any $n$ can be written as $6k$ or $6k+1$ or... or $6k+5$.  So
$$\eqalign{
  \sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}
  &=\sum_{k=0}^\infty\frac{\sin(12k+1)\frac\pi6}{(12k+1)^2}+\hbox{five more sums}\cr
  &=\sum_{k=0}^\infty\frac{\sin\frac\pi6}{(12k+1)^2}+\cdots\cr
  &=\frac12\sum_{k=0}^\infty\frac{1}{(12k+1)^2}+\cdots\ .\cr}$$
Then if we write
$$S_m=\sum_{k=0}^\infty\frac{1}{(12k+m)^2}$$
we have
$$G=S_1-S_3+S_5-S_7+S_9-S_{11}$$
and your sum $(2)$ is
$$\eqalign{
  -\frac12S_1&-S_3-\frac12S_5+\frac12S_7+S_9+\frac12S_{11}\cr
  &=-\frac12(S_1-S_3+S_5-S_7+S_9-S_{11})-\frac32S_3+\frac32S_9\cr
  &=-\frac12G-\frac32\frac19G\cr
  &=-\frac23G\ .\cr}$$
A: 
I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$. 

HINT:
In order to arrive at $(2)$, note that $\sin\left((2n+1)\frac\pi6\right)$ takes on the values
$$\sin\left((2n+1)\frac\pi6\right)=\begin{cases}\frac12 &,n=0,6,12,\dots\\\\
1 &,n=1,7,13,\dots\\\\
1/2&,n=2,8,14,\dots\\\\
-\frac12&,n=3,9,15,\cdots\\\\
-1&,n=4,10,16,\cdots\\\\
-\frac12&,n=5,11,17,\cdots
\end{cases}$$
