It's known that $SO(3)$ is isomorphic to SU(2), I am trying to establish a relation from exponentiation of their Lie Algebra, the formula I would like to prove is:

$$e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2}=R_{ij}(\hat{n},\theta)\sigma_{i} $$ Here is the prerequisite derivation, for $SO(3)$: The $SO(3)$ matrix in adjoint representation is given by: \begin{equation} (T_{i})_{jk}=-i\epsilon_{ijk} \end{equation} notice it satisfy the commutation relation: \begin{equation} [T_{i},T_{j}]_{\alpha \beta}=(T_{i})_{\alpha \gamma}(T_{j})_{\gamma \beta}-(T_{j})_{\alpha \gamma}(T_{i})_{\gamma \beta}=-\epsilon_{i\alpha \gamma}\epsilon_{j\gamma \beta}+\epsilon_{j\alpha \gamma}\epsilon_{i\gamma \beta} \end{equation} using the generalise Kroneckar delta: \begin{equation} [T_{i},T_{j}]_{\alpha \beta}=\epsilon_{\gamma i\alpha}\epsilon_{\gamma j \beta}-\epsilon_{\gamma j \alpha}\epsilon_{\gamma i \beta}=\delta_{j\beta}\delta_{i\alpha}-\delta_{i\beta}\delta_{j\beta}=i\epsilon_{ijk}(T_{k})_{\alpha \beta} \end{equation} given that $n$ is a unit vector, notice the square satisfy: \begin{equation} (n\cdot T)_{jm}^{2}=(n_{i}T_{i})_{jk}(n_{x}T_{x})_{kl}=-n_{i}n_{x}\epsilon_{kij}\epsilon_{klx}=\delta_{jl}-n_{l}n_{j} \end{equation} \begin{equation} (n\cdot T)_{jm}^{3}=(n_{i}T_{i})_{jk}(n_{x}T_{x})_{kl}(n_{y}T_{y})_{lm}=n_{i}n_{x}n_{y}(-i^{3})\epsilon_{kij}\epsilon_{klx}\epsilon_{lmy} \end{equation} using the Kroneckar delta: \begin{equation} (n\cdot T)_{jm}^{3}=in_{i}n_{x}n_{y}\epsilon_{lmy}(\delta_{il}\delta_{jx}-\delta_{ix}\delta_{jl})=-in_{x}^{2}n_{y}\epsilon_{jmy}=|n|^{2}(n\cdot T)_{jm} \end{equation} the eigenvalue of $(n\cdot T)$ then equivalent to: \begin{equation} (n\cdot T)[(n\cdot T)^{2}-1]=0 \;\;,\;\; n\cdot T=-1,0,1\;\;,\;\; J=1 \end{equation} using exponential to obtain the rotation: \begin{equation} R_{ij}(\hat{n},\theta);=\exp(-i\theta \hat{n}\cdot T)_{ij}=\sum_{n=0}^{\infty}\frac{(-i\theta \hat{n}\cdot T)_{ij}^{n}}{n!} \end{equation} expanding few terms to observe and grouping like terms: \begin{equation} I_{ij}-(\hat{n}\cdot T)_{ij}^{2}+(\hat{n}\cdot T)_{ij}^{2}(1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}+\dots)-i(\hat{n}\cdot T)_{ij}(\theta-\frac{\theta^{3}}{3!}+\dots) \end{equation} therefore the rotation is: \begin{equation} R_{ij}(\hat{n},\theta)=[1-(\hat{n}\cdot T)^{2}]_{ij}+(\hat{n}\cdot T)_{ij}^{2}\cos\theta-i(\hat{n}\cdot T)_{ij}\sin\theta \end{equation} further simplify in compact form: \begin{equation} R_{ij}(\hat{n},\theta)=[\delta_{ij}-(\delta_{ij}-n_{i}n_{j})]+(\delta_{ij}-n_{i}n_{j})\cos\theta-i(-i\epsilon_{kij}n_{k})\sin\theta \end{equation} which yield the Rodrigues Formula: \begin{equation} R_{ij}(\hat{n},\theta)=n_{i}n_{j}+(\delta_{ij}-n_{i}n_{j})\cos\theta-\epsilon_{ijk}n_{k}\sin\theta \end{equation} now also I obtained the $SU(2)$ The expression: \begin{equation} e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2} \end{equation} using the BCH formula: \begin{equation} \sigma_{j}+(-\frac{i\theta n_{i}}{2})[\sigma_{i},\sigma_{j}]+\frac{(-\frac{i\theta}{2}n_{i})^{2}}{2!}[\sigma_{i},[\sigma_{i},\sigma_{j}]]+\frac{((-\frac{i\theta n_{i}}{2}))^{3}}{3!}[\sigma_{i},[\sigma_{i},[\sigma_{i},\sigma_{j}]]]+\dots \end{equation} corresponding commutators are \begin{equation} [\sigma_{i},\sigma_{j}]=i\epsilon_{ijk}\sigma_{k} \;\;,\;\; i\epsilon_{ijk} [\sigma_{i},\sigma_{k}]=-\epsilon_{ijk}\epsilon_{ikj}\sigma_{j}=\sigma_{j} \end{equation} observing the pattern and grouping like terms \begin{equation} \sigma_{j}[1-\frac{1}{2!}(\frac{\theta}{2})^{2}+\dots]+i\epsilon_{ijk}\sigma_{k}n_{i}(-i)[\frac{1}{1!}(\frac{\theta}{2})-\frac{1}{3!}(\frac{\theta}{2})^{3}+\dots] \end{equation} using the Taylor expansion of sine and cosine: \begin{equation} e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2}=\sigma_{j}\cos\theta+\epsilon_{ijk}n_{i}\sigma_{k}\sin\theta \end{equation}

Nonetheless, I am not able to prove or disprove the first formula.

It's known that $SO(3)$ is (locally) isomorphic to $SU(2)$, I am trying to establish a relation from exponentiation of their Lie Algebra, the formula I would like to prove is:

$$e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2}=R_{ij}(\hat{n},\theta)\sigma_{i} $$ Here is the prerequisite derivation, for $SO(3)$: The $SO(3)$ matrix in adjoint representation is given by: \begin{equation} (T_{i})_{jk}=-i\epsilon_{ijk} \end{equation} notice it satisfy the commutation relation: \begin{equation} [T_{i},T_{j}]_{\alpha \beta}=(T_{i})_{\alpha \gamma}(T_{j})_{\gamma \beta}-(T_{j})_{\alpha \gamma}(T_{i})_{\gamma \beta}=-\epsilon_{i\alpha \gamma}\epsilon_{j\gamma \beta}+\epsilon_{j\alpha \gamma}\epsilon_{i\gamma \beta} \end{equation} using the generalise Kroneckar delta: \begin{equation} [T_{i},T_{j}]_{\alpha \beta}=\epsilon_{\gamma i\alpha}\epsilon_{\gamma j \beta}-\epsilon_{\gamma j \alpha}\epsilon_{\gamma i \beta}=\delta_{j\beta}\delta_{i\alpha}-\delta_{i\beta}\delta_{j\beta}=i\epsilon_{ijk}(T_{k})_{\alpha \beta} \end{equation} given that $n$ is a unit vector, notice the square satisfy: \begin{equation} (n\cdot T)_{jm}^{2}=(n_{i}T_{i})_{jk}(n_{x}T_{x})_{kl}=-n_{i}n_{x}\epsilon_{kij}\epsilon_{klx}=\delta_{jl}-n_{l}n_{j} \end{equation} \begin{equation} (n\cdot T)_{jm}^{3}=(n_{i}T_{i})_{jk}(n_{x}T_{x})_{kl}(n_{y}T_{y})_{lm}=n_{i}n_{x}n_{y}(-i^{3})\epsilon_{kij}\epsilon_{klx}\epsilon_{lmy} \end{equation} using the Kroneckar delta: \begin{equation} (n\cdot T)_{jm}^{3}=in_{i}n_{x}n_{y}\epsilon_{lmy}(\delta_{il}\delta_{jx}-\delta_{ix}\delta_{jl})=-in_{x}^{2}n_{y}\epsilon_{jmy}=|n|^{2}(n\cdot T)_{jm} \end{equation} the eigenvalue of $(n\cdot T)$ then equivalent to: \begin{equation} (n\cdot T)[(n\cdot T)^{2}-1]=0 \;\;,\;\; n\cdot T=-1,0,1\;\;,\;\; J=1 \end{equation} using exponential to obtain the rotation: \begin{equation} R_{ij}(\hat{n},\theta);=\exp(-i\theta \hat{n}\cdot T)_{ij}=\sum_{n=0}^{\infty}\frac{(-i\theta \hat{n}\cdot T)_{ij}^{n}}{n!} \end{equation} expanding few terms to observe and grouping like terms: \begin{equation} I_{ij}-(\hat{n}\cdot T)_{ij}^{2}+(\hat{n}\cdot T)_{ij}^{2}(1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}+\dots)-i(\hat{n}\cdot T)_{ij}(\theta-\frac{\theta^{3}}{3!}+\dots) \end{equation} therefore the rotation is: \begin{equation} R_{ij}(\hat{n},\theta)=[1-(\hat{n}\cdot T)^{2}]_{ij}+(\hat{n}\cdot T)_{ij}^{2}\cos\theta-i(\hat{n}\cdot T)_{ij}\sin\theta \end{equation} further simplify in compact form: \begin{equation} R_{ij}(\hat{n},\theta)=[\delta_{ij}-(\delta_{ij}-n_{i}n_{j})]+(\delta_{ij}-n_{i}n_{j})\cos\theta-i(-i\epsilon_{kij}n_{k})\sin\theta \end{equation} which yield the Rodrigues Formula: \begin{equation} R_{ij}(\hat{n},\theta)=n_{i}n_{j}+(\delta_{ij}-n_{i}n_{j})\cos\theta-\epsilon_{ijk}n_{k}\sin\theta \end{equation} now also I obtained the $SU(2)$ The expression: \begin{equation} e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2} \end{equation} using the BCH formula: \begin{equation} \sigma_{j}+(-\frac{i\theta n_{i}}{2})[\sigma_{i},\sigma_{j}]+\frac{(-\frac{i\theta}{2}n_{i})^{2}}{2!}[\sigma_{i},[\sigma_{i},\sigma_{j}]]+\frac{((-\frac{i\theta n_{i}}{2}))^{3}}{3!}[\sigma_{i},[\sigma_{i},[\sigma_{i},\sigma_{j}]]]+\dots \end{equation} corresponding commutators are \begin{equation} [\sigma_{i},\sigma_{j}]=i\epsilon_{ijk}\sigma_{k} \;\;,\;\; i\epsilon_{ijk} [\sigma_{i},\sigma_{k}]=-\epsilon_{ijk}\epsilon_{ikj}\sigma_{j}=\sigma_{j} \end{equation} observing the pattern and grouping like terms \begin{equation} \sigma_{j}[1-\frac{1}{2!}(\frac{\theta}{2})^{2}+\dots]+i\epsilon_{ijk}\sigma_{k}n_{i}(-i)[\frac{1}{1!}(\frac{\theta}{2})-\frac{1}{3!}(\frac{\theta}{2})^{3}+\dots] \end{equation} using the Taylor expansion of sine and cosine: \begin{equation} e^{-i\theta\hat{n}\cdot\sigma/2}\sigma_{j}e^{i\theta\hat{n}\cdot\sigma/2}=\sigma_{j}\cos\theta+\epsilon_{ijk}n_{i}\sigma_{k}\sin\theta \end{equation}

Nonetheless, I am not able to prove or disprove the first formula.