Show that $$\cos\theta = \cos(60^\circ)\cos(25^{\circ})\cos(48^{\circ})\cos(-122^{\circ})+\cos(60^{\circ})\sin(25^{\circ})\cos(48^{\circ})\sin(-122^{\circ})+\sin(60^{\circ})\sin(48^{\circ}) = 0.36299$$
The answer shows 0.36299 (68.716 degrees) but the answer I got was 0.704055146 I calculated it several times but still got the same wrong answer. What am I doing wrong? Yes--I did use a calculator. The calculator is in degrees, and yes--all angles are degrees.
$$\cos60^\circ\cos25^{\circ}\cos48^{\circ}\cos(-122^{\circ})+\cos60^{\circ}\sin25^{\circ}\cos48^{\circ}\sin(-122^{\circ})+\sin60^{\circ}\sin48^{\circ}=$$ $$=\cos60^{\circ}\cos48^{\circ}(\cos25^{\circ}\cos122^{\circ}-\sin25^{\circ}\sin122^{\circ})+\sin60^{\circ}\sin48^{\circ}=$$ $$=\cos60^{\circ}\cos48^{\circ}\cos147^{\circ}+\sin60^{\circ}\sin48^{\circ}$$ and $$\arccos\left(\cos60^{\circ}\cos48^{\circ}\cos147^{\circ}+\sin60^{\circ}\sin48^{\circ}\right)=68.7...^{\circ}.$$