How is it possible to force \$v(t)=\cos(t)\$ as voltage input to a fully discharged capacitor?
Remember that the capacitor stores energy in a electric field and \$i(t) = C \frac{dv(t)}{dt}\$.

If at \$t = 0, \ \cos(t) =1\$ then \$\frac{dv(t)}{dt} = \infty\$ so the current is not 0 A.

If however \$v(t) = \sin(\omega t)\$ then \$i(t) = C\frac{sin(t)}{dt} = \cos(t)\  \forall\ t\ge 0\$ then at \$t = 0,\ i(0)  = \cos(0) = 1\ A\$.

In simpler terms \$i(t) = c\frac{\Delta v(t)}{\Delta t}\$ and this means that you cannot change the voltage in the capacitor at \$t=0\$ from \$0\ V\$ to \$1\ V\$ at that instant.  Therefore \$v(t) = \sin(t)\$ is simpler to analyse from \$t=0\$. In the case where \$v(t) = \sin(t), i(t) = \cos(t)\$.