Actually, this is an exercise in Serre's Lectures on Mordell--Weil Theorem:
$K(A[n])$ always contains $\mu_n$ if $char(K)$ does not divide $n$ and $A$ is an abelian variety of positive dimension over $K$. (You don't need to assume that $K$ is a number field)
Here is a solution. First, it suffices to check the case when $n=\ell^m$ is a power of a prime $\ell$.
Second, if $A^t$ is the dual of $A$ then let us take a $K$-polarization $\lambda: A \to A^{t}$ of smallest possible degree. Then $\lambda$ is not divisible by $\ell$, i.e., $\ker(\lambda)$ does not contain the whole $A[\ell]$. (Otherwise, dividing $\lambda$ by $\ell$ we get a $K$-polarization of lesser degree.)
Then the image $\lambda(A[\ell^m])\subset A^t[\ell^m]$ contains a point of exact order $\ell^m$, say $Q$. Otherwise, $$\lambda(A[\ell^m])\subset A^t[\ell^{m-1}]$$ and therefore $A[\ell]=\ell^{m-1}A[\ell^m]$ lies in the kernel of $\lambda$, which is not the case.
Since $A[\ell^m]\subset A[K]$ and $\lambda$ is defined over $K$, the image $\lambda(A[\ell^m])$ lies in $A^t(K)$. In particular, $Q$ is a $K$-rational point on $A^t$.
Third, there is a nondegenerate Galois-equivariant Weil pairing $$e_n: A[\ell^m] \times A^t[\ell^m] \to \mu_{\ell^m}.$$ I claim that there is a point $P \in A[\ell^m]$ such that $e_n(P,Q)$ is a primitive $\ell^m$th root of unity. Indeed, otherwise $$e_n(A[\ell^m],Q) \subset \mu_{\ell^{m-1}}$$ and therefore nonzero $\ell^{m-1}Q$ is orthogonal to the whole $A[\ell^m]$ with respect to $e_n$, which contradicts the nondegeneracy of $e_n$.
So, $$\gamma:=e_n(P,Q)$$ is a primitive $\ell^m$th root of unity that lies in $K$, because both $P$ and $Q$ are $K$-points. Since cyclic $\mu_{\ell^m}$ is generated by $\gamma$, $$\mu_{\ell^m}\subset K.$$