Respuesta :

Yakoto

Your question has been heard loud and clear.

Lhs = 2cos[5A+3A/2]cos[5A-3A/2]+2cos[15A+7A/2]cos[15A-7A/2]

=2cos4AcosA+2cos11Acos4A

=2cos4A[cosA+cos11A]

=2cos4A[2cos[11A+A/2]Cos[11A-A/2]

=2cos4A2cos5Acos6A

=4cos4Acos5Acos6A=Rhs

Thank you