Реешение:
сos3acos7acos5a=cos5a*(cos10a+cos4a)/2=1/2(cos5a*cos10a+cos5acos4a)=
=1/2(1/2(cos15a+cos5a)+1/2(cos9a+cosa)=1/4(cosa+cos5a+cos9a+cos15a)
4cosa/2cosacos5a/2=4cosa(1/2(cos3a+cosa))=2cosa*cos3a+2cos^2a=cos2a+cos4a+2cos^2a=
=cos2a+cos4a+1+cos2a=2cos2a+cos4a+1.