= Solution
{c}
Write $T_{\theta,\mathbf b}(\mathbf v)=R_\theta\mathbf v+\mathbf b$. The <sine addition formula> and <cosine addition formula> give $R_\theta R_\psi=R_{\theta+\psi}$. Therefore <function composition> satisfies
$$
T_{\theta,\mathbf b}\circ T_{\psi,\mathbf c}=T_{\theta+\psi,\,R_\theta\mathbf c+\mathbf b}.
$$
The resulting <angle> is real and the new <translation> <vector> lies in $\mathbb R^2$, so $G$ is closed under <function composition>. <Associativity> holds because, for any <functions> $S,T,U$ and any $\mathbf v$, both $(S\circ T)\circ U$ and $S\circ(T\circ U)$ take $\mathbf v$ to $S(T(U(\mathbf v)))$.
The <identity element> is $T_{0,\mathbf0}$, since $R_0=I$. Every element has the two-sided <inverse element>
$$
\boxed{T_{\theta,\mathbf b}^{-1}=T_{-\theta,\,-R_{-\theta}\mathbf b}.}
$$
Indeed $R_\theta R_{-\theta}=R_{-\theta}R_\theta=I$, and substituting into the composition formula makes the <translation> <vector> zero in either order. These verify all the <group axioms>. Thus \b[$G$ is a <group> under <function composition>], the <special Euclidean group> in two dimensions.
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