Decomposition of tensor product space into direct sum Consider a tensor product space of two representations $j = \frac{3}{2}$ and $j = 1$. How to show $4 \otimes 3 = 6 \oplus 4 \oplus 2$?
 A: Since you know about $SU(2)$ characters, this is a doable exercise. Let me remind you that the character of spin-$j$ is 
$$\chi_j(t) = \sum_{m=-j}^j e^{2imt} = \frac{\sin (2j+1)t}{\sin t}.$$
The crucial property is that the character of a tensor product is the product of characters, i.e. $\chi_{j \otimes j'}(t) = \chi_j(t) \chi_{j'}(t).$ In your case, you're interested in $j=3/2$ and $j=1$ so we have
$$\chi_{3/2 \otimes 1}(t) = \frac{\sin 4t \sin 3t}{(\sin t)^2}.$$
At the same time, the characters are a complete basis of functions, so you can write
$$\chi_{3/2 \otimes 1}(t) = \sum_{j=0,1/2,\ldots} a_j \, \chi_j(t).$$
The coefficients $a_j$ encode the multiplicity of spin $j$ appearing in the tensor product $3/2 \otimes 1$. (Or in your language $4 \otimes 3$.)
You can compute the $a_j$ in several ways. Most physicists would use Mathematica to solve for the $a_j$ -- this is a completely rigorous approach. Alternatively, you can also the orthogonality of the characters i.e. one may write the multiplicities $a_j$ as
$$a_j = \frac{1}{\pi} \int_0^{2\pi}dt \sin^2(t) \, \chi_j(t) \, \chi_{3/2 \otimes 1}(t).$$
By manipulating the integrals you can show that $a_j = 1$ if $j=1/2,3/2,5/2$ and zero otherwise. This is a fully rigorous proof of the fact that $3/2 \otimes 1 = 5/2 \oplus 3/2 \oplus 1/2$.
