Formula for curvature of two intersecting surfaces in terms of their normal curvature. I have been privately reading DoCarmo recently, and have been attempting to do some of the problems. I am stuck on this one, it is problem 14 in section 3.2 for those interested. If someone could show me how to do it, or even offer a hint, it would be much appreciated. 
If the surface $S_1$ intersects the surface $S_2$ along the regular curve $C$, then the curvature $k$ of $C$ at $p\in C$ is given by
$k^2\sin^2\theta=\lambda_1^2+\lambda_2^2-2\lambda_1\lambda_2\cos\theta$
Where $\lambda_1$ and $\lambda_2$ are the normal curvatures at $p$, along th tangent line to $C$, of $S_1$ and $S_2$, respectively, and $\theta$ is the angle made up by the normal vectors in $S_1$ and $S_2$ at $p$.
My attempt at the proof:
Since $\lambda_i=\langle T',\nu_i\rangle$, where $T$ is the tangent vector along $C$ and $\nu_i$ is the vector normal to $S_i$ at $p$. So,
$\lambda_1^2+\lambda_2^2-2\lambda_1\lambda_2\cos\theta$
$=\langle T',\nu_1\rangle^2+\langle T',\nu_2\rangle^2-2\langle T',\nu_1\rangle\langle T',\nu_2\rangle\cos\theta$
$=|\langle T',\nu_1\rangle\nu_1-\langle T',\nu_2\rangle\nu_2|^2$
$=|\langle kN,\nu_1\rangle\nu_1-\langle kN,\nu_2\rangle\nu_2|^2$
$=k^2|\langle N,\nu_1\rangle\nu_1-\langle N,\nu_2\rangle\nu_2|^2$
So if I could show that $=|\langle N,\nu_1\rangle\nu_1-\langle N,\nu_2\rangle\nu_2|^2=\sin^2\theta$ I am done. Any ideas?
 A: It is important that all three vectors $N,\nu_1,\nu_2$ lie in the same plane (the normal plane to $C$). Let's take $N$ for the polar axis in this plane. Denote the polar angles of $\nu_1$ and $\nu_2$ by $\alpha$ and $\beta$, respectively. Since $\langle N,\nu_1\rangle = \cos\alpha $ and $\langle N,\nu_2\rangle = \cos\beta $, the desired identity takes the form
$$
\cos^2\alpha +\cos^2\beta - 2 \cos\alpha\cos\beta \cos (\alpha-\beta) = \sin^2(\alpha-\beta)
\tag1$$
There may be some clever way to prove (1), but I will simply use $\cos (\alpha-\beta)=\cos\alpha\cos\beta+\sin \alpha\sin\beta$, rewriting the left side of (1) as 
$$\begin{split}
\cos^2\alpha + \cos^2\beta &- 2 \cos^2\alpha\cos^2\beta - 2\cos\alpha\cos\beta\sin \alpha\sin\beta \\ &=\cos^2\alpha(1-\cos^2\beta)+\cos^2\beta(1-\cos^2\alpha)-2\cos\alpha\cos\beta\sin \alpha\sin\beta  \\& = \cos^2\alpha\sin^2\beta+\cos^2\beta \sin^2\alpha -2\cos\alpha\cos\beta\sin \alpha\sin\beta  
\\&=(\sin\alpha\cos\beta-\cos\alpha\sin\beta)^2 \\& = \sin^2(\alpha-\beta)
\end{split}$$ 
For a different proof, see Curvature of Regular curve at a point.
A: You can also use the formula
$$u \times (v \times w)= \langle u,w \rangle v - \langle u,v \rangle w,$$
together with the fact that the three vectors lie in the same plane.
