Using the fact that sin(A+B)=sinAcosB+cosAsinB and the differentiation, obtain the sum formula for cosines.
Consider A and B are function of x.
Differentiating both side w.r.t. x , we get
d(sin(A+B))dx=d(sinAcosB+cosAsinB)dx
d(sin(A+B))dx=d(sinA.cosB)dx+d(cosA.sinB)dx
cos(A+B).d(A+B)dx=d(sinA.cosB)dx+d(cosA.sinB)dx
Using Product Rule : (uv)′=u′v+v′u,
cos(A+B).(dAdx+dBdx)=(cosBd(sinA)dx+sinAd(cosB)dx)+(d(cosA)dx.sin B+d(sinB)dx.cosA)
cos(A+B).(dAdx+dBdx)=cosA.dAdx.cosB−sinB.dBdxsinA−sinA.dAdx.sinB+cosB.dBdx.cosA
cos(A+B).(dAdx+dBdx)=cosA.dAdx.cosB−sinA.dAdxsinB−sinB.dBdx.sinA+cosB.dBdx.cosA
cos(A+B).(dAdx+dBdx)=dAdx(cosAcosB−sinAsinB)+dBdx(−sinBsinA+cosBcosA)
cos(A+B).(dAdx+dBdx)=(cosAcosB−sinAsinB)(dAdx+dBdx)
Thus,
cos(A+B)=cosAcosB−sinAsinB
Hence, Proved.