Question

Arătați că:
sin 24° sin^2 72° + sin 36° sin^2 78° = cos 12° cos 18°

Answer 1

$\sin24^{\circ}\sin^272^{\circ} + \sin36^{\circ}\sin^278^{\circ} = \\\\ = \sin(2\cdot 12)^{\circ} \sin^2(90-18)^{\circ} + \sin (2\cdot 18)^{\circ}\sin^2 (90-12)^{\circ} = \\ \\ =2\sin12^{\circ}\cos12^{\circ}\cos^218^{\circ} +2\sin18^{\circ}\cos18^{\circ}\cos^212^{\circ} = \\ \\ = 2\cos12^{\circ}\cos18^{\circ}\Big(\sin12^{\circ}\cos18^{\circ} + \sin18^{\circ}\cos12^{\circ}\Big) = \\ \\ = 2\cos12^{\circ}\cos18^{\circ} \sin(12+18)^{\circ} =$

$= 2\cos12^{\circ}\cos18^{\circ}\sin30^{\circ} =\\ \\ = 2\cos 12^{\circ}\cos 18^{\circ}\cdot \dfrac{1}{2} = \\ \\ = \boxed{\cos12^{\circ} \cos 18^{\circ}}$