Question

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

(a) $\frac{\sqrt{5}+1}{8}$ (b) $\frac{\sqrt{5}-1}{8}$ (c) $\frac{\sqrt{5}+1}{5}$ (d) $\frac{\sqrt{5}+1}{2\sqrt{2}}$

Answer 1

$$\begin{gathered} {\cos }^{2}48°-{\sin }^{2}12° \\ =\cos \left(48°+12°\right)\cos \left(48°-12°\right)\left[\cos \left(A+B\right)\cos \left(A-B\right)={\cos }^{2}A-{\sin }^{2}B\right] \\ =\cos 60°\cos 36° \\ =\frac{1}{2}\times \left(\frac{\sqrt{5}+1}{4}\right) \\ =\frac{\sqrt{5}+1}{8} \end{gathered}$$

Hence, the correct answer is option A.