Step 1:  Read the question and make a note of the important points

              NOTE:  A, B and C are interior angles of right angle triangle ABC

                              then A + B + C = 180\degree

                              Prove: \sin \frac{B+C}{2} = \cos \frac{A}{2}

Step 2:  Take the given hint and manipulate it  

                HINT:  A + B + C = 180\degree

                          Dividing the above equation by 2 on both sides.

                          \Rightarrow    \frac{A +B +C}{2} = \frac{180\degree}{2}  

                          Separate the angles according to our requirement      

                             \Rightarrow  \frac{A}{2} + \frac{B +C}{2} = 90\degree   

                             \Rightarrow    \frac{B +C}{2} = 90\degree - \frac{A}{2}

                          Apply the \sin 0n both the sides

                         \Rightarrow   \sin (\frac{B +C}{2}) = \sin (90\degree - \frac{A}{2})

                                      \sin (\frac{B +C}{2}) = \cos (\frac{A}{2})

                                                         Hence, proved