Question:
The value of cos (36° − A) cos (36° + A) + cos (54° + A) cos (54° − A) is
(a) sin 2A
(b) cos 2A
(c) cos 3A
(d) sin 3A
Solution:
(b) cos 2A
$\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\cos \left(54^{\circ}+A\right) \cos \left(54^{\circ}-A\right)$
$=\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\sin \left[90^{\circ}-\left(54^{\circ}+A\right)\right] \sin \left[90^{\circ}-\left(54^{\circ}-A\right)\right] \quad\left[\right.$ Since $\left.\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
$=\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\sin \left(36^{\circ}-A\right) \sin \left(36^{\circ}+A\right)$
$=\cos \left(36^{\circ}+A-36^{\circ}+A\right) \quad[$ Using $\cos (\mathrm{A}-\mathrm{B})$ formula $]$
$=\cos 2 A$