If $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$, then $\cos ^2 48^{\circ}-\sin ^2 12^{\circ}$ has the value

Question

Aicharya · Accepted Answer

$\begin{array}{ll} & \cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B}) \ 	herefore \quad & \cos ^2 48^{\circ}-\sin ^2 12^{\circ}=\cos \left(60^{\circ}ight) \cdot \cos \left(36^{\circ}ight) \ & =\frac{1}{2} \cdot\left(1-2 \sin ^2 \frac{36}{2}ight) \ & =\frac{1}{2}\left(1-2 \sin ^2 18^{\circ}ight) \ & =\frac{1}{2}\left[1-2\left(\frac{\sqrt{5}-1}{4}ight)^2ight] \ & =\frac{\sqrt{5}+1}{8}\end{array}$

If $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$, then $\cos ^2 48^{\circ}-\sin ^2 12^{\circ}$ has the value

Solution