If A, B and C are interior angles of triangle ABC, then show that \sin \frac{B + C}{2} = \cos \frac{A}{2}

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