Question

cos(A+B)/2=sin(C/2)​

Answer 1

Step-by-step explanation:

In triangle ABC,

A + B + C = 180°

$\therefore \: A + B = 180° - C \\ \frac{A + B}{2} = \frac{ 180° - C}{2} \\ \frac{A + B}{2} = 90 \degree - \frac{C}{2} \\ cos(\frac{A + B}{2}) =cos( 90 \degree - \frac{C}{2}) \\ cos(\frac{A + B}{2}) = \: sin\frac{C}{2} \\ thus \: proved$