The value of cos

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 $]$

 

$=\cos 2 A$

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