We were taught that $\ce{Cr_2O_7^{2-}}$ (dichromate) in $\ce{H_2SO_4}$ gives $\ce{CrO_3}$
$$\ce{Cr_2O_7^{2-} +H_2SO_4 -> CrO_3 + ...}$$
Link: wikipedia
But we know that dichromate in acidic medium converts to $\rm Cr(III)$.
$$\ce{ Cr_2O^{2−}_7 + 14 H3O+ + 6 e− → 2 Cr^{3+} + 7 H2O}$$
Link: wikipedia
How is this possible?
Both statements on the behaviour of $\ce{Cr2O7^{2-}}$ are true, but you are confusing two different reactions.
The formation of chromium trioxide from (sodium) dichromate in sulfuric acd is not a redox reaction. There is no partner that could be oxidized by $\ce{Cr2O7^{2-}}$. Concentrated sulfuric acid is an oxidant itself and metals like copper are dissolved (oxidized to $\ce{Cu^{2+}}$) while not hydrogen is generated.
Remember also that sulfuric acid is a strong, oxidizing and dehydrating acid. This effect is used in the production of $\ce{CrO3}$, which can be considered the anhydride of chromic acid ($\ce{H2CrO4}$). Note that chromium in $\ce{Cr2O7^{2-}}$, $\ce{H2CrO4}$, and $\ce{CrO3}$ always has the same oxidation state.
The second reaction that you mentioned does happen too, and this is indeed part of a redox reaction in which $\ce{Cr(VI)}$ is reduced to $\ce{Cr(III)}$. However, this can only happen when a suitable reaction partner is present, which in turn is oxidized by $\ce{Cr2O7^{2-}}$.
