Consider the alternating group $\mathcal A_n$ ($n$ is an odd integer). Do $(12\cdots n)$ and $(12)(34)$ generate $\mathcal A_n$? In other words, $\langle (12\cdots n),(12)(34)\rangle =\mathcal A_n$.
I know that $\langle (12\cdots n),(123)\rangle=\mathcal A_n$. But how can we make $(123)$?
Collecting the bits from comments to an answer.
By conjugating $\sigma_1=(12)(34)$ by powers of $(123\cdots n)$ we get the permutations $\sigma_i=(i\ i+1)(i+2\ i+3)$, where we count modulo $n$, if some of the entries are $>n$. Then, using the assumption $2\nmid n$ we get $$ \tau=\sigma_1\sigma_3\sigma_5\cdots\sigma_{n-2}=(12)(34)(34)(56)\cdots(n-2\ n-1)(n\ 1)=(12)(n1)=(1n2) $$ after a mass cancellation. (I am interpreting the products in the order that the rightmost one is the first to be applied, if you use the other possible convention, then you get the inverse of this result).
So $\tau^{-1}=(n12)$. Conjugating this by $(123\cdots n)$ gives $(123)$ taking us to the point, where the OP can use the known result.