cos 48^{circ} cdot cos 12^{circ} = ? | vedclass

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Question

(D) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.$\cos 48^{\circ} \cdot \cos 12^{\circ} = \frac{1}{2} [2 \cos 48^{\circ} \cos 12^{\cir

Vedclass · Accepted Answer

$\frac{3+\sqrt{5}}{8}$

Vedclass · Answer

(D) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.$\cos 48^{\circ} \cdot \cos 12^{\circ} = \frac{1}{2} [2 \cos 48^{\circ} \cos 12^{\circ}]$$= \frac{1}{2} [\cos(48^{\circ}+12^{\circ}) + \cos(48^{\circ}-12^{\circ})]$$= \frac{1}{2} [\cos 60^{\circ} + \cos 36^{\circ}]$Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$,$= \frac{1}{2} [\frac{1}{2} + \frac{\sqrt{5}+1}{4}]$$= \frac{1}{2} [\frac{2 + \sqrt{5} + 1}{4}]$$= \frac{3+\sqrt{5}}{8}$

$\cos 48^{\circ} \cdot \cos 12^{\circ} = ?$

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