$\operatorname{Cos}^2 48^{\circ}-\sin ^2 12^{\circ}=$
,if $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$

Question

$\operatorname{Cos}^2 48^{\circ}-\sin ^2 12^{\circ}=$ ,if $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$

Aicharya · Accepted Answer

$\begin{aligned} & \cos ^2 48^{\circ}-\sin ^2 12^{\circ} \ & =\cos \left(48^{\circ}+12^{\circ}ight) \cdot \cos \left(48^{\circ}-12^{\circ}ight) \ & =\cos 60^{\circ} \cdot \cos 36^{\circ} \ & =\frac{1}{2}\left\{1-2 \sin ^2 18^{\circ}ight\} \ & =\frac{1}{2}\left\{1-2 	imes\left(\frac{\sqrt{5}-1}{4}ight)^2ight\} \ & =\frac{1}{2} 	imes\left\{1-2 	imes \frac{5+1-2 \sqrt{5}}{16}ight\} \ & =\frac{1}{2}-\frac{6-2 \sqrt{5}}{16}=\frac{1+\sqrt{5}}{8}\end{aligned}$

$\operatorname{Cos}^2 48^{\circ}-\sin ^2 12^{\circ}=$ ,if $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$

Solution