If $\mathrm{x}=\mathrm{a}\left(\mathrm{\theta }+\mathrm{sin\theta }\right), \mathrm{y}=\mathrm{a}\left(1-\mathrm{cos\theta }\right),$ then ${\mathrm{y}}_2$ at $\mathrm{\theta }=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ is

$$\begin{gathered} \mathrm{x}=\mathrm{a}(\mathrm{\theta }+\mathrm{sin\theta }) \\ \frac{\mathrm{dx}}{\mathrm{d\theta }}=\mathrm{a}(1+\mathrm{cos\theta }) \\ \mathrm{y}=\mathrm{a}(1-\mathrm{cos\theta }) \\ \frac{\mathrm{dy}}{\mathrm{d\theta }}=\mathrm{a}\left(\mathrm{sin\theta }\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{dy}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{d\theta }$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{dx}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{d\theta }$}\right.}}=\frac{\mathrm{a} \mathrm{sin\theta }}{\mathrm{a}(1+\mathrm{cos\theta })} \\ =\frac{2\sin {\displaystyle \frac{\mathrm{\theta }}{2}} \cos {\displaystyle \frac{\mathrm{\theta }}{2}}}{2{\cos }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}=\tan \frac{\mathrm{\theta }}{2} \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\sec }^2 \frac{\mathrm{\theta }}{2}·\frac{1}{2}·\frac{\mathrm{d\theta }}{\mathrm{dx}} \\ =\frac{1}{2} {\sec }^2\frac{\mathrm{\theta }}{2}·2\frac{1}{2\mathrm{a} {\cos }^2{\displaystyle \frac{\mathrm{\theta }}{2}}} \\ =\frac{1}{4\mathrm{a}}-\frac{1}{{\cos }^4{\displaystyle \frac{\mathrm{\theta }}{2}}}=\frac{1}{\mathrm{a}{\left(2{\cos }^2{\displaystyle \frac{\mathrm{\theta }}{2}}\right)}^2} \\ =\frac{1}{\mathrm{a}{(1+\mathrm{cos\theta })}^2}=\frac{1}{\mathrm{a}{\left(1+{\displaystyle \frac{1}{\sqrt{2}}}\right)}^2} \\ =\frac{2}{2{\left(\sqrt{2}+1\right)}^2}=\frac{2}{\mathrm{a}(3+2\sqrt{2})} \\ =\frac{2(3-2\sqrt{2})}{\mathrm{a}} \end{gathered}$$