Question 10.39: Prove that (a) sn(z + 2K) = -sn z, (b) cn(z + 2K) = -cn z.
Prove that (a) \operatorname{sn}(z+2 K)=-\operatorname{sn} z, (b) \operatorname{cn}(z+2 K)=-\operatorname{cn} z.
We have z=\int_{0}^{\phi} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}} so that \phi=\operatorname{am} z and \sin \phi=\operatorname{sn} z, \cos \phi=\operatorname{cn} z. Now
\begin{aligned} \int\limits_{0}^{\phi+\pi} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}} & =\int\limits_{0}^{\pi} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}+\int\limits_{\pi}^{\phi+\pi} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}} \\ & =2 \int\limits_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}+\int\limits_{0}^{\phi} \frac{d \psi}{\sqrt{1-k^{2} \sin ^{2} \psi}} \end{aligned}
using the transformation \theta=\pi+\psi. Hence, \phi+\pi= am (z+2 K).
Thus we have
(a) \operatorname{sn}(z+2 K)=\sin \{\operatorname{am}(z+2 K)\}=\sin (\phi+\pi)=-\sin \phi=- an z
(b) \operatorname{cn}(z+2 K)=\cos \{\operatorname{am}(z+2 K)\}=\cos (\phi+\pi)=-\cos \phi=-\mathrm{cn} z