 Solution

For a time-dependent type-two generating function for a canonical transformation, the differential identitygives $K=H+{\partial }_t{F}_2$, with the partial derivative taken at fixed $(q,P)$. Here $F_2=\cos q\sin P$ has no explicit time dependence, so the Hamiltonian correction isAs always, adding a function of time alone to the new Hamiltonian leaves Hamilton's equations unchanged. Choosing a time-independent type-two generating function for a canonical transformation fixes that irrelevant freedom and gives the displayed $\Lambda =0$.