Evaluate cos^2 48^{circ} - sin^2 12^{circ},gi | vedclass

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Question

(D) Using the identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$:$\cos^2 48^{\circ} - \sin^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cos(48^{\ci

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Using the identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$:$\cos^2 48^{\circ} - \sin^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cos(48^{\circ} - 12^{\circ})$$= \cos 60^{\circ} \cos 36^{\circ}$Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = 1 - 2\sin^2 18^{\circ}$:$= \frac{1}{2} (1 - 2\sin^2 18^{\circ})$$= \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}$

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

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