This is related to Hartshorne's Exercise III.2.7 here one shows that $H^1(S^1,\mathscr{R})=0$. Here, $\mathscr{R}$ is the sheaf of germs of continuous real-valued functions. The method of proof is to look at the long exact sequence and show that we only need surjectivity of $\Gamma(X,\mathscr{F})\rightarrow\Gamma(X,\mathscr{F}/\mathscr{R})$ where $\mathscr{F}$ is the sheaf of all real-valued functions on $S^1$.
On examining the proof, only two properties seemed to be used. (1) $S^1$ is compact and (2) one can require that the finite cover $\{U_i\}$ satisfiy $U_i\cap U_j\cap U_k=\emptyset$.
My immediate thought is that we can claim (using the same proof) that $H^1(D^n,\mathscr{R})=0$ and $H^1(S^n,\mathscr{R})=0$ for $n\geq 1$. Can someone confirm if this is indeed the case (I think these sheaves should be soft)?