cos ^2 48^{circ}-sin ^2 12^{circ} = . . . . | vedclass

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Question

(C) We use the identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$.Here,$A = 48^{\circ}$ and $B = 12^{\circ}$.$\cos^2 48^{\circ} - \sin^2 12^{\circ}

Vedclass · Accepted Answer

$\frac{\sqrt{5}+1}{8}$

Vedclass · Answer

(C) We use the identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$.Here,$A = 48^{\circ}$ and $B = 12^{\circ}$.$\cos^2 48^{\circ} - \sin^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cos(48^{\circ} - 12^{\circ})$$= \cos 60^{\circ} \cos 36^{\circ}$We know that $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.Therefore,the expression equals $\frac{1}{2} 	imes \frac{\sqrt{5}+1}{4} = \frac{\sqrt{5}+1}{8}$.

$\cos ^2 48^{\circ}-\sin ^2 12^{\circ} = $ . . . . . . ,if $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$

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