Added Note 1: Derivation of the rotation 
matrix
For those who would like to see the details, we work out here the general rotation matrix for a system with spin (total angular momentum) $j$. It is really not very important to work out the general case; once you have the idea, you can find the general results in tables in many books. On the other hand, after coming this far you might like to see that you can indeed understand even the very complicated formulas of quantum mechanics, such as Eq. (18.35), that come into the description of angular momentum.  We extend the arguments of Section 18-4 to a system with spin $j$, which we consider to be made up of $2j$ spin one-half objects. The state with $m=j$ would be $\ket{+\,+\,+\dotsb+}$ (with $2j$ plus signs). For $m=j-1$, there will be $2j$ terms like $\ket{+\,+\dotsb+\,+\,-}$, $\ket{+\,+\dotsb+\,-\,+}$, and so on. Let’s consider the general case in which there are $r$ plusses and $s$ minuses—with $r+s=2j$. Under a rotation about the $z$-axis each of the $r$ plusses will contribute $e^{+i\phi/2}$. The result is a phase change of $(r/2-s/2)\phi$. You see that \begin{equation} \label{Eq:III:18:59} m=\frac{r-s}{2}. \end{equation} Just as for $j=\tfrac{3}{2}$, each state of definite $m$ must be the linear combination with plus signs of all the states with the same $r$ and $s$—that is, states corresponding to every possible arrangement which has $r$ plusses and $s$ minuses. We assume that you can figure out that there are $(r+s)!/r!s!$ such arrangements. To normalize each state, we should divide the sum by the square root of this number. We can write \begin{align} \biggl[\frac{(r+s)!}{r!s!}\biggr]^{-1/2}\kern{-3.5ex} \{\ket{\underbrace{+\,+\,+\dotsb+\,+}_r\, \underbrace{-\,-\,-\dotsb-\,-}_s}&\notag\\[1ex] +\,(\text{all rearrangements of order})\}&\notag\\[1.5ex] \label{Eq:III:18:60} =\ket{j,m}& \end{align} with \begin{equation} \label{Eq:III:18:61} j=\frac{r+s}{2},\quad m=\frac{r-s}{2}. \end{equation}  It will help our work if we now go to still another notation. Once we have defined the states by Eq. (18.60), the two numbers $r$ and $s$ define a state just as well as $j$ and $m$. It will help us keep track of things if we write \begin{equation} \label{Eq:III:18:62} \ket{j,m}=\ket{\tover{r}{s}}, \end{equation} where, using the equalities of (18.61) \begin{equation*} r=j+m,\quad s=j-m. \end{equation*} Next, we would like to write Eq. (18.60) with a new special notation as \begin{equation} \label{Eq:III:18:63} \ket{j,m}=\ket{\tover{r}{s}}= \biggl[\frac{(r+s)!}{r!s!}\biggr]^{+1/2}\kern{-3.5ex} \{\ket{+}^r\ket{-}^s\}_{\text{perm}}. \end{equation} Note that we have changed the exponent of the factor in front to plus $\tfrac{1}{2}$. We do that because there are just $N=(r+s)!/r!s!$ terms inside the curly brackets. Comparing (18.63) with (18.60) it is clear that \begin{equation*} \{\ket{+}^r\ket{-}^s\}_{\text{perm}} \end{equation*} is just a shorthand way of writing \begin{equation*} \frac{\{\ket{+\,+\dotsb-\,-}+\text{all rearrangements}\}}{N}, \end{equation*} where $N$ is the number of different terms in the bracket. The reason that this notation is convenient is that each time we make a rotation, all of the plus signs contribute the same factor, so we get this factor to the $r$th power. Similarly, all together the $s$ minus terms contribute a factor to the $s$th power no matter what the sequence of the terms is.  Now suppose we rotate our system by the angle $\theta$ about the $y$-axis. What we want is $R_y(\theta)\,\ket{\tover{r}{s}}$. When $R_y(\theta)$ operates on each $\ket{+}$ it gives \begin{equation} \label{Eq:III:18:64} R_y(\theta)\,\ket{+}=\ket{+}C+\ket{-}S, \end{equation} where $C=\cos\theta/2$ and $S=-\sin\theta/2$. When $R_y(\theta)$ operates on each $\ket{-}$ it gives \begin{equation*} R_y(\theta)\,\ket{-}=\ket{-}C-\ket{+}S. \end{equation*} So what we want is \begin{equation} \begin{aligned} &R_y(\theta)\ket{\tover{r}{s}}\\[.5ex] &=\biggl[\frac{(r\!+\!s)!}{r!s!}\biggr]^{1/2}\kern{-1ex} R_y(\theta)\,\{\ket{+}^r\ket{-}^s\}_{\text{perm}}\\[1ex] &=\biggl[\frac{(r\!+\!s)!}{r!s!}\biggr]^{1/2}\kern{-1ex} \{(R_y(\theta)\,\ket{+})^r(R_y(\theta)\,\ket{-})^s\}_{\text{perm}}\\[1ex] &=\biggl[\frac{(r\!+\!s)!}{r!s!}\biggr]^{1/2}\kern{-1ex} \{(\ket{+}C\!+\!\ket{-}S)^r(\ket{-}C\!-\!\ket{+}S)^s\}_{\text{perm}}. \end{aligned} \label{Eq:III:18:65} \end{equation} Now each binomial has to be expanded out to its appropriate power and the two expressions multiplied together. There will be terms with $\ket{+}$ to all powers from zero to $(r+s)$. Let’s look at all of the terms which have $\ket{+}$ to the $r'$ power. They will appear always multiplied with $\ket{-}$ to the $s'$ power, where $s'=2j-r'$. Suppose we collect all such terms. For each permutation they will have some numerical coefficient involving the factors of the binomial expansion as well as the factors $C$ and $S$. Suppose we call that factor $A_{r'}$. Then Eq. (18.65) will look like \begin{equation} \label{Eq:III:18:66} R_y(\theta)\,\ket{\tover{r}{s}}=\sum_{r'=0}^{r+s} \{A_{r'}\,\ket{+}^{r'}\ket{-}^{s'}\}_{\text{perm}}. \end{equation} Now let’s say that we divide $A_{r'}$ by the factor $[(r'+s')!/r'!s'!]^{1/2}$ and call the quotient $B_{r'}$. Equation (18.66) is then equivalent to \begin{equation} \label{Eq:III:18:67} R_y(\theta)\,\ket{\tover{r}{s}}\!=\!\sum_{r'=0}^{r+s} B_{r'}\!\biggl[\frac{(r'\!+\!s')!}{r'!s'!}\biggr]^{1/2}\kern{-2.5ex} \{\ket{+}^{r'}\ket{-}^{s'}\}_{\text{perm}}. \end{equation} (We could just say that this equation defines $B_{r'}$ by the requirement that (18.67) gives the same expression that appears in (18.65).)  With this definition of $B_{r'}$ the remaining factors on the right-hand side of Eq. (18.67) are just the states $\ket{\tover{r'}{s'}}$. So we have that \begin{equation} \label{Eq:III:18:68} R_y(\theta)\,\ket{\tover{r}{s}}=\sum_{r'=0}^{r+s} B_{r'}\,\ket{\tover{r'}{s'}}, \end{equation} with $s'$ always equal to $r+s-r'$. This means, of course, that the coefficients $B_{r'}$ are just the matrix elements we want, namely \begin{equation} \label{Eq:III:18:69} \bracket{\tover{r'}{s'}}{R_y(\theta)}{\tover{r}{s}}=B_{r'}. \end{equation}  Now we just have to push through the algebra to find the various $B_{r'}$. Comparing (18.65) with (18.67)—and remembering that $r'+s'=r+s$—we see that $B_{r'}$ is just the coefficient of $a^{r'}b^{s'}$ in the following expression: \begin{equation} \label{Eq:III:18:70} \biggl(\frac{r'!s'!}{r!s!}\biggr)^{1/2} (aC+bS)^r(bC-aS)^s. \end{equation} It is now only a dirty job to make the expansions by the binomial theorem, and collect the terms with the given power of $a$ and $b$. If you work it all out, you find that the coefficient of $a^{r'}b^{s'}$ in (18.70) is  \begin{equation} \label{Eq:III:18:71} \biggl[\frac{r'!s'!}{r!s!}\biggr]^{1/2}\sum_k (-1)^kS^{r-r'+2k}C^{s+r'-2k}\cdot \frac{r!}{(r-r'+k)!(r'-k)!}\cdot \frac{s!}{(s-k)!k!}. \end{equation} 
  \begin{align} \biggl[\frac{r'!s'!}{r!s!}\biggr]^{1/2}\kern{-2.5ex}&\lower .3ex {\sum_k (-1)^kS^{r-r'\!+2k}C^{s+r'\!-2k}}\notag\\[-2ex] \label{Eq:III:18:71} &\quad\;\times\frac{r!}{(r\!-\!r'\!+\!k)!(r'\!-\!k)!}\!\cdot\! \frac{s!}{(s\!-\!k)!k!}. \end{align}   The sum is to be taken over all integers $k$ which give terms of zero or greater in the factorials. This expression is then the matrix element we wanted.  Finally, we can return to our original notation in terms of $j$, $m$, and $m'$ using  \begin{equation*} r=j+m,\quad r'=j+m',\quad s=j-m,\quad s'=j-m'. \end{equation*} 
 \begin{alignat*}{2} &r&=&\,j+m,\quad &r'&=&\,j+m',\\[.5ex] &s&=&\,j-m,\quad &s'&=&\,j-m'. \end{alignat*}  Making these substitutions, we get Eq. (18.35) in Section 18-4. 