Question

If A, B, C are the interior angles of a ΔABC, show that `cos[(B+C)/2] = sin A/2`

Answer 1

We have to prove: `cos[(B+C)/2] = sin A/2`

Since we know that in triangle ABC

`A + B + C = 180^@`

`=> B + C = 180^@ - A`

Dividing by 2 on both sides, we get

`=> (B +C)/2 = 90^@ - A/2`

`=> cos (B + C)/2 = cos (90^@ - A/2)`

`=> cos  (B + C)/2 = sin A/2`

Proved