Question

(sinA+sinB)^2+(cosA+cosB)^2=4cos^2 (A-B/2)

Answer 1

sinA^2+sinB^2+2*sinA*sinB+cosA^2+cosB^2+2*cosA*cosB=
= sinA^2+cosA^2+sinB^2+cosB^2 + 2(sinA*sinB+cosA*cosB)
= 1 + 1 +2*cos(A-B)
=2 + 2*cos(2*$\frac{A-B}{2}$)
=2+ 2(2 cos^2($\frac{A-B}{2}$)-1)
=2+ 4 cos^2($\frac{A-B}{2}$) -2
=4 cos^2($\frac{A-B}{2}$)