Let $\pi_n$ be the irreduicble representations of $SU(2)$. Let $n\geq m $ I want to show that $\pi_m \otimes \pi_n =\pi_{n+m}\oplus\pi_{n+m-2}\ldots \pi_{n-m} $.
I know that $\chi_{\pi_n}=\frac{\sin(n-1)\phi}{\sin(\phi)}$. In generally for characters we have $\chi_{\rho \otimes \pi}=\chi_\rho \chi_\pi$ and for a compact group $\pi \sim \bigoplus_{\rho\in \hat{G}}\left\langle \chi_\pi,\chi_\rho\right\rangle\rho$ where $\hat{G}$ is the irreducible representations of $G$.
This gives that $\pi_m \otimes \pi_n=\bigoplus_{\rho\in \hat{G}}\left\langle \chi_\rho,\chi_{\pi_m}\chi_{\pi_m}\right\rangle\rho=\bigoplus_{\rho\in \hat{G}}\left\langle \chi_\rho,\frac{\sin(m-1)\phi}{\sin(\phi)}\frac{\sin(n-1)\phi}{\sin(\phi)}\right\rangle\rho$.
I think its also true that the only irreducible representations in $SU(2)$ are $\pi_n$. Which gives $\bigoplus_{j\in \mathbb N}\left\langle \frac{\sin(j-1)\phi}{\sin(\phi)},\frac{\sin(m-1)\phi}{\sin(\phi)}\frac{\sin(n-1)\phi}{\sin(\phi)}\right\rangle\pi_j$.
I do not know how to calculate $\left\langle \frac{\sin(j-1)\phi}{\sin(\phi)},\frac{\sin(m-1)\phi}{\sin(\phi)}\frac{\sin(n-1)\phi}{\sin(\phi)}\right\rangle$, and i'm not actually sure how to define the inner product.
Update: Another approach I have been trying is that two representations are isomorphic if and only if their characters are the same. The character of $\pi_{n+m}\oplus\pi_{n+m-2}\ldots \pi_{n-m} $ is given by $\chi_{\pi_{n+m}}+\chi_{\pi_{n+m-1}}+...\chi_{\pi_{n-m}}$. The character of $\chi_{\pi_n\otimes \pi_m}=\chi_{\pi_n}\chi_{\pi_m}$. But I am having trouble showing these are equall.
