Question

The value of $\cos \left(36°-A\right)\cos \left(36°+A\right)+\cos \left(54°-A\right)\cos \left(54°+A\right)$ is
(a) cos 2A
(b) sin 2A
(c) cos A
(d) 0

Answer 1

(a) cos 2A

$$\begin{gathered} \cos \left(36°-\mathrm{A}\right)\cos \left(36°+\mathrm{A}\right)+\cos \left(54°-\mathrm{A}\right)\cos \left(54°+\mathrm{A}\right) \\ =\cos \left[90°-\left(54°+\mathrm{A}\right)\right]\cos \left[90°-\left(54°-\mathrm{A}\right)\right]+\cos \left(54°-\mathrm{A}\right)\cos \left(54°+\mathrm{A}\right) \\ =\sin \left(54°+\mathrm{A}\right)\sin \left(54°-\mathrm{A}\right)+\cos \left(54°-\mathrm{A}\right)\cos \left(54°+\mathrm{A}\right)\left[\cos \left(90°-\theta \right)=\sin \theta \right] \\ =\cos \left(54°+\mathrm{A}-54°+\mathrm{A}\right)\left[\cos \left(\mathrm{A}-\mathrm{B}\right)=\mathrm{cosAcosB}+\mathrm{sinAsinB}\right] \\ =\cos 2\mathrm{A} \end{gathered}$$