Question

(cosα + cosβ)^2 + (sinα − sinβ)^2 = 4cos^2[α + β]/2

Answer 1

lhs=cos^2α+cos^2β+2cosα*cosβ+sin^2α+sin^2β-2sinα*sinβ

now sin^2x+cos^2x=1
also
2cosA*cosB=cos(A+B)+cos(A-B)
2sinA*sinB=cos(A-b)-cos(A+B)
so
(sin^2α+cos^2α)+(sin^2β+cos^2β)+cos(α+β)+cos(α-β)+cos(α-β)-cos(α+β)
2+2cos(α-β)

now
cos2x=2cos^2x-1
sooooooooooo
2+2(2cos^2(α-β)/2-1)
4cos^2(α-β)/2