Question

(cosA-cosB)2 + (sinA-sinB)2 = 4sin2(A+B/2)

Answer 1

it's a good answer for the students

Answer 2

Answer:

Consider LHS: $(\cos A-\cos B)^2 + (\sin A-\sin B)^2$

$=\cos^2 A+\cos^2 B-2\cos A\cos B+\sin^2 A+\sin^2B-2\sin A\sin B\\\\=\cos^2 A+\sin^2 A+\sin^2B+\cos^2 B-2\cos A\cos B-2\sin A\sin B..........[\cos^2x+\sin^2x=1]\\\\=1+1-2(\cos A\cos B+\sin A\sin B)\\\\=2-2(\sin(A+B))\\\\=2(1-\cos(A+B))\\\\=2(2\sin^2(\frac{A+B}{2}))........[1-\cos2x=2\sin^2x]\\\\=4\sin^2(\frac{A+B}{2})$....RHS.

∴$(\cos A-\cos B)^2 + (\sin A-\sin B)^2=4\sin^2(\frac{A+B}{2})$   =RHS.

Hence proved.