If sin 18^{circ} = frac{sqrt{5}-1}{4},then th | vedclass

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Question

(C) We use the trigonometric identity: $\cos ^2 A - \sin ^2 B = \cos(A+B) \cdot \cos(A-B)$.Applying this to the given expression with $A = 48^{\circ}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We use the trigonometric identity: $\cos ^2 A - \sin ^2 B = \cos(A+B) \cdot \cos(A-B)$.Applying this to the given expression with $A = 48^{\circ}$ and $B = 12^{\circ}$:$\cos ^2 48^{\circ} - \sin ^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cdot \cos(48^{\circ} - 12^{\circ})$$= \cos(60^{\circ}) \cdot \cos(36^{\circ})$Since $\cos(60^{\circ}) = \frac{1}{2}$ and $\cos(36^{\circ}) = 1 - 2\sin ^2(18^{\circ})$:$= \frac{1}{2} \cdot (1 - 2\sin ^2 18^{\circ})$$= \frac{1}{2} \left[ 1 - 2 \left( \frac{\sqrt{5}-1}{4} \right)^2 \right]$$= \frac{1}{2} \left[ 1 - 2 \left( \frac{5 + 1 - 2\sqrt{5}}{16} \right) \right]$$= \frac{1}{2} \left[ 1 - \frac{6 - 2\sqrt{5}}{8} \right] = \frac{1}{2} \left[ \frac{8 - 6 + 2\sqrt{5}}{8} \right]$$= \frac{1}{2} \left[ \frac{2 + 2\sqrt{5}}{8} \right] = \frac{1 + \sqrt{5}}{8}$.

If $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,then the value of $\cos ^2 48^{\circ} - \sin ^2 12^{\circ}$ is:

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