I think you mean: cos(60-A)×cos(60+A)×cos(A)=¼cos(3A).
cos(60-A)=cos(A)/2+sin(A)√3/2;
cos(60+A)=cos(A)/2-sin(A)√3/2;
cos(60-A)cos(60+A)=cos2(A)/4-¾sin2(A).
cos(60-A)cos(60+A)cos(A)=cos3(A)/4-¾sin2(A)cos(A)=
cos3(A)/4-¾cos(A)+¾cos3(A)=cos3(A)-¾cos(A).
cos(3A)=cos(2A)cos(A)-sin(2A)sin(A)=
2cos3(A)-cos(A)-2sin2(A)cos(A)=
2cos3(A)-cos(A)-2(1-cos2(A))cos(A)=
2cos3(A)-cos(A)-2cos(A)+2cos3(A)=
4cos3(A)-3cos(A)=4(cos3(A)-¾cos(A)).
Therefore cos(3A)/4=cos(60-A)cos(60+A)cos(A) QED