The value of cos
Question:

The value of $\cos \left(36^{\circ}-A\right) \cos \left(36^{\circ}+A\right)+\cos \left(54^{\circ}-A\right) \cos \left(54^{\circ}+A\right)$ is

(a) cos 2A

(b) sin 2A

(c) cos A

(d) 0

Solution:

(a) cos 2A

$\cos \left(36^{\circ}-\mathrm{A}\right) \cos \left(36^{\circ}+\mathrm{A}\right)+\cos \left(54^{\circ}-\mathrm{A}\right) \cos \left(54^{\circ}+\mathrm{A}\right)$

$=\cos \left[90^{\circ}-\left(54^{\circ}+\mathrm{A}\right)\right] \cos \left[90^{\circ}-\left(54^{\circ}-\mathrm{A}\right)\right]+\cos \left(54^{\circ}-\mathrm{A}\right) \cos \left(54^{\circ}+\mathrm{A}\right)$

$=\sin \left(54^{\circ}+\mathrm{A}\right) \sin \left(54^{\circ}-\mathrm{A}\right)+\cos \left(54^{\circ}-\mathrm{A}\right) \cos \left(54^{\circ}+\mathrm{A}\right) \quad\left[\cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$

$=\cos \left(54^{\circ}+\mathrm{A}-54^{\circ}+\mathrm{A}\right) \quad[\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}]$

$=\cos 2 \mathrm{~A}$

 

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