Let $t\mapsto \beta(t)\in S^2 \subset\mathbb{R}^3$ be a regular curve. We define $$t\mapsto \gamma(t):=\int_{t_0}^t \beta(t)\times \beta'(t)dt.$$ Show that $\gamma$ has a constant torsion.
We have that \begin{align*} \gamma'&=\beta \times \beta',\\ \gamma''&=\beta \times \beta'',\\ \gamma'''&=\beta' \times \beta''+\beta \times \beta''' \end{align*} We can calculate torsion with the following formula $$\tau=\dfrac{|\gamma',\gamma'',\gamma'''|}{|\gamma'\times \gamma''|^2}=\dfrac{|\beta \times \beta',\beta \times \beta'',\beta' \times \beta''+\beta \times \beta'''|}{|(\beta \times \beta')\times (\beta \times \beta'')|^2}$$ I know that $$(\beta \times \beta')\times (\beta \times \beta'')=|\beta,\beta',\beta''|\beta-|\beta,\beta',\beta|\beta''=|\beta,\beta',\beta''|\beta$$ But what now?
Note that $\beta \times \beta'$, $\beta \times \beta''$ and $\beta \times \beta'''$ are linear dependent (all three are orthogonal to $\beta$), hence $$ \det(\beta \times \beta', \beta \times \beta'',\beta \times \beta''') = 0 $$ For the other summand in the numerator, due to \begin{align*} \det(a,b,c \times d) &= a \cdot \bigl(b \times (c\times d)\bigr)\\ &= a \cdot \bigl((b\cdot d)c - (b \cdot c)d\bigr)\\ &= (b\cdot d)(a \cdot c) - (b \cdot c)(a \cdot d) \end{align*} We have \begin{align*} \det(\beta \times \beta', \beta \times \beta'', \beta' \times \beta'') &= \bigl((\beta \times \beta'') \cdot \beta'\bigr) \bigl((\beta \times \beta') \cdot \beta'' \bigr) \\ &= \det(\beta', \beta, \beta'') \det(\beta'', \beta, \beta')\\ &= -\det(\beta, \beta', \beta'')^2 \end{align*} As the denominator equals $\det(\beta, \beta', \beta'')^2$ (note that $|\beta| = 1$), the quotient is constant.