Question

(sinα+sinβ)²+(cosα+cosβ)²=4cos²((α-β)/2)

Answer 1

Sin²a+sin²b+2sinasinb+cos²a+cos²b+2cosacosb=
=(sin²a+cos²a)+(sin²b+cos²b)+2(sinasinb+cosacosb)=
=1+1+2cos(a-b)=2(1+cos(a-b))

4cos²((α-β)/2)=4*(1+cos(a-b))/2=2(1+cos(a-b))