Find the Value of cos²48° – sin²12°
Question:
\[ \cos^2 48^\circ-\sin^2 12^\circ \]Solution
Use the identity
\[ \sin^2\theta=\cos^2(90^\circ-\theta) \]Therefore,
\[ \sin^2 12^\circ=\cos^2 78^\circ \]Hence,
\[ \cos^2 48^\circ-\sin^2 12^\circ = \cos^2 48^\circ-\cos^2 78^\circ \]Using
\[ \cos^2 A=\frac{1+\cos 2A}{2} \] \[ =\frac{1+\cos96^\circ}{2} -\frac{1+\cos156^\circ}{2} \] \[ =\frac{\cos96^\circ-\cos156^\circ}{2} \]Using the identity
\[ \cos C-\cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \] \[ =\frac{-2\sin126^\circ\sin(-30^\circ)}{2} \] \[ =\sin126^\circ \] \[ =\sin54^\circ \]Answer
\[ \boxed{\sin54^\circ} \]Since
\[ \sin54^\circ=\cos36^\circ =\frac{1+\sqrt5}{4}, \]the numerical value is
\[ \boxed{\frac{1+\sqrt5}{4}} \]