f(z)=sin(z)

=sin(x+iy)

= sin(x) cos(iy) +cos(x) sin(iy) ------------(1)

Recognizing that in general, cos(y) =[ e^(iy)+ e^(-iy) ]/2,

then cos(iy) = { e^[i(iy)]+ e^-[i(iy)] } /2 (simply replace y by iy on the RHS)

= [e^(-y) +e^(y)]/2 = cosh(y)

Similarly, recognizing that in general, sin(y) =[ e^(iy)- e^(-iy) ]/ (2i) ,

then sin(iy) ={ e^[i(iy)] - e^-[i(iy)] } /(2i)

= [e^(-y) - e^(y)]/ (2i)

= (i)* [e^(y) - e^(-y)]/2 = i sinh(y)

Substituting these into (1), we have f(z)= sin(x) cosh(y) + i cos(x) sinh(y) ,

where u(x,y) = sin(x) cosh(y) and v(x,y) =cos(x) sinh(y) (shown)

Hope this helps. Good luck for your upcoming test. Peace.