Derivation of geodesic on $m$-sphere $\gamma : \mathbb{R} \rightarrow S^m$ is given by $\gamma(t) = \cos(t |v|)p \ + \frac{\sin(t |v|)}{|v|} v$ Given $p \in S^m$ and $0 \neq v \in T_pS^m = p^{\perp}$, the geodesic $\gamma : \mathbb{R} \rightarrow S^m$ seems to take the form by $\gamma(t) = \cos(t |v|)p \ + \frac{\sin(t |v|)}{|v|} v$, where $t \in \mathbb{R}$.
This is pretty much a given as fact in all differential geometry books I've seen so far. I assume this is some kind of supposedly trivial parameterization obtained by intersecting $S^m$ with the plane spanned by $\big\{p, \frac{v}{|v|} \big\}$, but I haven't been able to connect that with the actual closed form of the geodesic $\gamma$.
I'd tremendously appreciate a derivation of this result, as I haven't been able to find this anywhere else.
 A: Consider $\mathbb{S}^m \subset \mathbb{R}^{m+1}$ with the induced metric. What is to be known is that for a riemannian submanifold $(N,g,\nabla) \subset (M,\bar g,\bar \nabla)$, the Levi-Civita connection $\nabla$ is the orthogonal projection of the ambiant Levi-Civita connexion $\bar \nabla$ on the tangent space. Thus, if $c$ is a parametrized curve in $\mathbb{S}^m$, $\nabla_{c'} c' = \left(\bar \nabla_{c'} c'\right)^{\perp}$. Using this, one can show that the curve
\begin{align}
\gamma (t) = \cos(t\|v\|)p +  \sin(t\|v\|)\frac{v}{\|v\|}
\end{align}
is a geodesic of $\mathbb{S}^m$. Indeed, in the ambiant space $\mathbb{R}^{m+1}$, the connexion is trivial, so the computation shows that $\bar\nabla_{\gamma'(t)}\gamma'(t) = \gamma''(t)=-\|v\|\sin(t\|v\|)p + \cos(t\|v\|)v$ is orthogonal to $T_p\mathbb{S}^m$, and $\nabla_{\gamma'}\gamma' = 0$. Thus, $\gamma$ is a geodesic.
To conclude, by uniqueness of geodesics (because $\nabla_{\gamma'}\gamma'$ is a linear second order differential equation in charts), if $c$ is a geodesic with $c(0)=p$ and $c'(0) = v$, $c = \gamma$.
Edit Here is another proof, more constructive. The idea is to show that a geodesic has to lie inside a linear plane of $\mathbb{R}^{m+1}$. For that, let $p \in \mathbb{S}^m$ and $v \in T_p\mathbb{S}^m$ be a non-zero tangent vector. Let $P = \mathrm{span}(p,v)$. Consider the linear isometry $u$ of $\mathbb{R}^{m+1}$ that is the reflexion with respect to $P$. One can show it is an isometry of $\mathbb{S}^m$ because it stabilizes the sphere and preserves the scalar product on tangent spaces.
Now that we know $u$ is an isometry of $\mathbb{S}^m$, let $\gamma$ be the geodesic of $\mathbb{S}^m$ with $\gamma(0) = p$ and $\gamma'(0)=v$. Then $u\circ \gamma$ is also a geodesic (easy computation) with the same initial data. Thus, $u\circ \gamma = \gamma$ and $\gamma$ has to lie in the set of points in $\mathbb{S}^m$ unvariant by $u$, that is, $P \cap \mathbb{S}^m$. So $\gamma$ is a curve lying insinde the great circle $\mathbb{S}^m\cap P$. As it is a circle and as a geodesic has to be parametrized with constant velocity, is has to be of the form given above.
