
theorem Th3:
  for a, b being Real holds sin(a-b) = sin(a)*cos(b)-cos(a)*
  sin(b) & cos(a-b) = cos(a)*cos(b)+sin(a)*sin(b)
proof
  let th1, th2 be Real;
  thus sin(th1-th2)=sin.(th1+(-th2)) by SIN_COS:def 17
    .=(sin.(th1)) *(cos.(-th2))+(cos.(th1)) * (sin.(-th2)) by SIN_COS:74
    .=(sin.(th1)) *(cos.(th2))+(cos.(th1)) * (sin.(-th2)) by SIN_COS:30
    .=(sin.(th1)) *(cos.(th2))+(cos.(th1)) * -(sin.(th2)) by SIN_COS:30
    .=(sin(th1)) *(cos.(th2))-(cos.(th1)) * (sin.(th2)) by SIN_COS:def 17
    .=(sin(th1)) *(cos(th2))-(cos.(th1)) * (sin.(th2)) by SIN_COS:def 19
    .=(sin(th1)) *(cos(th2))-(cos(th1)) * (sin.(th2)) by SIN_COS:def 19
    .=(sin(th1)) *(cos(th2))-(cos(th1)) * (sin(th2)) by SIN_COS:def 17;
  thus cos(th1-th2)=cos.(th1+(-th2)) by SIN_COS:def 19
    .=(cos.(th1)) *(cos.(-th2))-(sin.(th1)) * (sin.(-th2)) by SIN_COS:74
    .=(cos.(th1)) *(cos.(th2))-(sin.(th1)) * (sin.(-th2)) by SIN_COS:30
    .=(cos.(th1)) *(cos.(th2))-(sin.(th1)) * -(sin.(th2)) by SIN_COS:30
    .=(cos(th1)) *(cos.(th2))+(sin.(th1)) * (sin.(th2)) by SIN_COS:def 19
    .=(cos(th1)) *(cos(th2))+(sin.(th1)) * (sin.(th2)) by SIN_COS:def 19
    .=(cos(th1)) *(cos(th2))+(sin(th1)) * (sin.(th2)) by SIN_COS:def 17
    .=(cos(th1)) *(cos(th2))+(sin(th1)) * (sin(th2)) by SIN_COS:def 17;
end;
