cos(36^{circ}-A) cos(36^{circ}+A) + cos(54^{c | vedclass

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(A) Using the identity $\cos(x)\cos(y) + \sin(x)\sin(y) = \cos(x-y)$ and the co-function identity $\cos(90^{\circ}-	heta) = \sin(	heta)$:Given expre

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$\cos(2A)$

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(A) Using the identity $\cos(x)\cos(y) + \sin(x)\sin(y) = \cos(x-y)$ and the co-function identity $\cos(90^{\circ}-	heta) = \sin(	heta)$:Given expression: $\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \cos(54^{\circ}+A) \cos(54^{\circ}-A)$Since $54^{\circ}+A = 90^{\circ}-(36^{\circ}-A)$ and $54^{\circ}-A = 90^{\circ}-(36^{\circ}+A)$,we have:$\cos(54^{\circ}+A) = \sin(36^{\circ}-A)$ and $\cos(54^{\circ}-A) = \sin(36^{\circ}+A)$Substituting these into the expression:$\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \sin(36^{\circ}-A) \sin(36^{\circ}+A)$This is in the form $\cos(x)\cos(y) + \sin(x)\sin(y)$ where $x = 36^{\circ}-A$ and $y = 36^{\circ}+A$.$= \cos((36^{\circ}-A) - (36^{\circ}+A))$$= \cos(36^{\circ}-A-36^{\circ}-A)$$= \cos(-2A)$Since $\cos(-	heta) = \cos(	heta)$,the result is $\cos(2A)$.

$\cos(36^{\circ}-A) \cos(36^{\circ}+A) + \cos(54^{\circ}+A) \cos(54^{\circ}-A) = $

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