Why $\cos(df_p(v_1),df_p(v_2))=\cos(v_1,v_2)\Rightarrow|df_p(v)|^2=\lambda(p)^2 |v|^2$?

Question

On the 169th page of do Carmo's Riemannian Geometry, the author state that $$ \cos (df_p(v_1), df_p(v_2)) = \cos (v_1,v_2) \Rightarrow |df_p(v)|^2=\lambda(p)^2 |v|^2 $$ where $v,v_1,v_2$ are the vectors at $p$, and $(v_1,v_2)$ is the angle of $v_1,v_2$. And $f:U \rightarrow \mathbb R^n$, where $U$ is an open subset of $\mathbb R^n$.

Answer 1

I think this relates to a result in an exercise from Do Carmo's earlier book on curves and surfaces, which asks the reader to prove that a differentiable map $\varphi: S_1 \to S_2$ (where $S_1$ and $S_2$ are regular surfaces in $\mathbb R^3$) preserves angles if and only if $\varphi$ is locally conformal. I don't think it's too difficult, and the answer given here: Proof that angle-preserving map is conformal gives a good summary of how to go about it.

Anyway, Do Carmo defines a conformal map as a diffeomorphism $\varphi: S \to \bar{S}$ such that for all $p \in S$ and all $v_1, v_2 \in T_p(S)$ we have $$\langle d \varphi_p(v_1), d \varphi_p(v_2) \rangle_{\varphi(p)} = \lambda(p)^2 \langle v_1, v_2 \rangle_{p},$$ where $\lambda^2$ is a nowhere-zero differentiable function on $S$, so I think it just follows from this.

You may need to adapt this result for smooth manifolds (rather than surfaces in $\mathbb R^3$), but I'm pretty sure this is the basic story.