Question

In ΔABC, prove that `sin  (A + B)/2 = cos  C/2`.

Answer 1

Given: In triangle ABC, A + B + C = 180°.

To Prove: `sin  (A + B)/2 = cos  C/2`.

Proof [Step-wise]:

1. Start from the triangle angle-sum:

A + B + C = 180°

2. Rearrange to express (A + B):

A + B = 180° – C

3. Divide both sides by 2:

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

4. Use the co-function identity sin (90° – x) = cos x.

Applying it with `x = C/2` gives `sin  (A + B)/2`

= `sin (90^circ - C/2)`

= `cos  C/2`

Hence proved: `sin  (A + B)/2 = cos  C/2`.