Question

in triangle ABC prove. cos(A+B/2) = sin c/2​

Answer 1

Answer:

Firstly,A+B+C=180°

A+B=180°-C

divide both sides by 2

A+B/2=180-C/2

Multiply both sides by cos

cos(A+B/2)=cos(180°-C/2)

cos(A+B/2)=cos(90°-C/2)

cos(A+B/2)=sin(c/2) because cos(90°-A)=sin(A)