The value of cos
Question:

The value of $\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}$ is

(a) $\frac{\sqrt{5}+1}{8}$

(b) $\frac{\sqrt{5}-1}{8}$

(c) $\frac{\sqrt{5}+1}{5}$

(d) $\frac{\sqrt{5}+1}{2 \sqrt{2}}$

Solution:

$\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}$

$=\cos \left(48^{\circ}+12^{\circ}\right) \cos \left(48^{\circ}-12^{\circ}\right) \quad\left[\cos (A+B) \cos (A-B)=\cos ^{2} A-\sin ^{2} B\right]$

$=\cos 60^{\circ} \cos 36^{\circ}$

$=\frac{1}{2} \times\left(\frac{\sqrt{5}+1}{4}\right)$

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

Hence, the correct answer is option A.

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