Let $\alpha:I\to {\mathbb R}^3$ be a cylindrical helix with a unit vector $u$ such that $u\cdot T_{\alpha}$ is a constant for all $t\in I$. For $t_0\in I$, the curve $$\gamma(t)=\alpha(t)-((\alpha(t)-\alpha(t_0))\cdot u)u$$ is called a cross-sectional curve of the cyliner on which $\alpha$ lies. Prove that the curvature of $\gamma$ is $\frac{\kappa_{\alpha}}{sin^2\theta}$, where $\kappa_{\alpha}$ is the curvature of $\alpha$, and $\theta$ is the angle between $T_{\alpha}$ and $u$
I've proved that:
$$\sin\theta=\frac{\kappa_{\alpha}}{\sqrt{{\kappa_{\alpha}}^2+\tau_{\alpha}^2}}$$
so we're trying to show that:
$$\kappa_{\gamma}=\kappa_{\alpha}+\frac{\tau_{\alpha}^2}{\kappa_{\alpha}}$$
Expressing both sides in terms of their derivatives, we have:
$$\frac{|\gamma'\times\gamma''|}{|\gamma'|^3}=\frac{|\alpha'\times\alpha''|}{|\alpha'|^3}+\frac{|\alpha'|^3}{|\alpha'\times\alpha''|}\frac{((\alpha'\times\alpha")\cdot{\alpha'''})^2}{|\alpha'\times\alpha"|^4}$$.
Since $\gamma'=\alpha'-(\alpha'\cdot u)u$ and $\gamma''=\alpha''-(\alpha''\cdot u)u$, we get:
$$\frac{|(\alpha'-(\alpha'\cdot u)u)\times{\alpha''}-(\alpha''\cdot u)\alpha'\times u|}{|\alpha'-(\alpha'\cdot u)u|^3}=\frac{|\alpha'\times\alpha''|}{|\alpha'|^3}+\frac{|\alpha'|^3}{|\alpha'\times\alpha''|}\frac{((\alpha'\times\alpha")\cdot{\alpha'''})^2}{|\alpha'\times\alpha"|^4}$$
Because of the complexity of the equation, I think I should approach to it at some other perspective. I also believe that the conclusion $\sin \theta=\frac{\kappa_{\alpha}}{\sqrt{{\kappa_{\alpha}}^2+\tau_{\alpha}^2}}$ should still be utilized. I hope someone could give me a clue.