Докажете тъждеството

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Докажете тъждеството

[S.B.]

[tex]\sin \alpha \sin(60 ^\circ - \alpha)\sin(60 ^\circ + \alpha ) =[/tex] [tex]\frac{1}{4} \sin 3 \alpha[/tex]

[tex]\sin \alpha \sin(60 ^\circ - \alpha )\sin (60 ^\circ + \alpha) =[/tex]

[tex]= \sin \alpha [\sin 60 ^\circ\cos \alpha - \cos 60 ^\circ\sin \alpha][\sin 60 ^\circ\cos \alpha + \cos 60 ^\circ \sin \alpha] =[/tex]

[tex]= \sin \alpha.[ \frac{ \sqrt{3} }{2}.\cos \alpha - \frac{1}{2}.\sin \alpha].[ \frac{ \sqrt{3} }{2}.\cos \alpha + \frac{1}{2}.\sin \alpha] =[/tex]

[tex]= \sin \alpha ( \frac{3}{4} \cos^{2 } \alpha - \frac{1}{4} \sin^{2 } \alpha ) =[/tex]

[tex]= \frac{1}{4}\sin \alpha (3 \cos^{2 } \alpha - \sin^{2 } \alpha) =[/tex]

[tex]= \frac{1}{4}\sin \alpha [ 3.(1 - \sin^{2 } \alpha) - \sin^{2 } \alpha] =[/tex]

[tex]= \frac{1}{4}\sin \alpha (3 - 3.\ sin^{2 } \alpha - \sin^{2 } \alpha) =[/tex]

[tex]= \frac{1}{4}\sin \alpha(3 - 4 \sin^{2 } \alpha ) =[/tex]

[tex]= \frac{1}{4}(3.\sin \alpha - 4. \sin^{3 } \alpha ) =[/tex]

[tex]= \frac{1}{4}\sin 3 \alpha[/tex]
$$\Rightarrow \sin \alpha \sin(60 ^\circ - \alpha)\sin(60 ^\circ + \alpha ) = \frac{1}{4}\sin 3 \alpha $$