cos(u+v)cos(v) + sin(u+v)sin(v) = cos(u) Remembering the identity cos(A-B) = cos(A)cos(B) + sin(A)sin(B) we recognize that the left side, cos(u+v)cos(v) + sin(u+v)sin(v) is simply cos(A)cos(B) + sin(A)sin(B) with u+v substituted for A and v substituted for B. we have cos[(u+v)-v] cos(u+v-v) cos(u) Edwin