Derivative of Delta sine

Theorem

The following formula holds: $$\sin_p^{\Delta}(t,t_0)=p(t)\cos_p(t,t_0),$$ where $\sin_p$ denotes the $\Delta$-$\sin_p$ function and $\cos_p$ denotes the $\Delta$-$\cos_p$ function.

Proof

Compute $$\begin{array}{ll} \sin^{\Delta}_p(t,t_0) &= \dfrac{1}{2i} \dfrac{\Delta}{\Delta t} \left( e_{ip}(t,t_0) - e_{-ip}(t,t_0) \right) \\ &= \dfrac{ip}{2i} ( e_{ip}(t,t_0) + e_{-ip}(t,t_0) ) \\ &= \dfrac{1}{2} (e_{ip}(t,t_0)+e_{-ip}(t,t_0)) \\ &= p\cos_p(t,t_0), \end{array}$$ as was to be shown. █