Question

Prove That Cos(36o - A)Cos(36o + A) + Cos (54o + A)Cos(54o - A) = Cos2A

Answer 1

Cos (36 - A) (Cos (36+A) + Cos (54+A) COS(54-A)

=  1/2 * [ cos 72 + Cos 2A ] + 1/2 [ Cos 108 + Cos 2A ]
 = Cos 2A + 1/2 [ cos 72 + cos 108 ]
 = cos 2 A + 1/2 [ cos 72 + Cos (180-72) ]
 = cos 2 A + 1/2 [ cos 72 - cos 72 ]
= cos 2A

Answer 2

You have, $cos(54^{o}+A)=sin(90^{o}-(54^{o}+A))=sin(36^{o}-A)$ and similarly $cos(54^{o}-A)=sin(90^{o}-(54^{o}-A))=sin(36^{o}+A)$.
Then the expression becomes $cos(36^{o}-A)cos(36{o}+A)+sin(36^{o}-A)sin(36^{o}+A)$.
This is $cos((36^{o}-A)-(36^{o}+A))=cos(-2A)=cos(2A)$.