(cos^2(37)-sin^2(23))/sin104Решите пожалуйста

Решите задачу:

$\frac{\cos^2 37^{\circ} - \sin^2 23^{\circ}}{\sin 104^{\circ}} = \frac{\cos^2 37^{\circ} - \sin^2 23^{\circ}}{\sin 104^{\circ}} = \frac{(\cos 37^{\circ} - \sin 23^{\circ})(\cos 37^{\circ} + \sin 23^{\circ})}{\sin 104^{\circ}} = \\\\ = \frac{(\sin 53^{\circ} - \sin 23^{\circ})(\sin 53^{\circ} + \sin 23^{\circ})}{\sin 104^{\circ}} = \frac{2\sin 15^{\circ}\cos 38^{\circ}*2\sin38^{\circ}\cos 15^{\circ}}{2\sin 52^{\circ}\cos 52^{\circ}} =$

$= \frac{4\sin 15^{\circ}\cos 15^{\circ}\cos 38^{\circ}\sin38^{\circ}}{2\sin(90^{\circ} - 38^{\circ})\cos(90^{\circ} - 38^{\circ})} = \frac{4\sin 15^{\circ}\cos 15^{\circ}\cos 38^{\circ}\sin38^{\circ}}{2\cos 38^{\circ}\sin38^{\circ}} = \\\\ = 2\sin 15^{\circ}\cos 15^{\circ} = \sin 30^{\circ} = \frac{1}{2}$

$\boxed{ \ \sin a \pm \sin b = 2\sin \frac{a \pm b}{2}\cos\frac{a \mp b}{2} \ }$