(cosα)^3×cos3α(sinα)^3×sin3α
=(1/16)(sin2a)^3*sin6a
=(1/64)(3sin2a-sin6a)sin6a
=(1/128)[3(cos4a-cos8a)-(1-cos12a)]
=(1/128)[3cos4a-3cos8a+cos12a-1].