${cos}^{2} 48^{o} - {sin}^{2} 12^{o}$ is equal to

\frac{\sqrt{5} - 1}{4}

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

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\frac{\sqrt{3} - 1}{4}

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\frac{\sqrt{3} + 1}{2 \sqrt{2}}

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Solution

The correct option is B $\frac{\sqrt{5} + 1}{8}$
Given ${cos}^{2} 48 - {sin}^{2} 12$

we know that ${cos}^{2} x - {cos}^{2} y = cos (x + y) cos (x - y)$

$= cos (48 - 12) cos (48 + 12)$

$= cos 36 cos 60 = (\frac{\sqrt{5} + 1}{4}) (\frac{1}{2}) = \frac{\sqrt{5} + 1}{8}$

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