Krishna
0

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