cos^2 48^circ - sin^2 12^circ = | vedclass

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Question

(B) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cdot \cos(A - B)$.Substituting $A = 48^\circ$ and $B = 12^\circ$:$\cos^2 48

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cdot \cos(A - B)$.Substituting $A = 48^\circ$ and $B = 12^\circ$:$\cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ + 12^\circ) \cdot \cos(48^\circ - 12^\circ)$$= \cos(60^\circ) \cdot \cos(36^\circ)$We know that $\cos(60^\circ) = \frac{1}{2}$ and $\cos(36^\circ) = \frac{\sqrt{5} + 1}{4}$.Therefore,the expression becomes $\frac{1}{2} \cdot \left( \frac{\sqrt{5} + 1}{4} \right) = \frac{\sqrt{5} + 1}{8}$.

$\cos^2 48^\circ - \sin^2 12^\circ = $

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