Odpowiedź:
Wiemy, że :
[tex]sin(\alpha -\beta )=sin\alpha cos\beta -cos\alpha sin\beta[/tex]
Oraz [tex]\beta \in (\pi ,\frac{3}{2} \pi )[/tex]
Wtedy :
[tex]sin(\beta -\frac{1}{3} \pi )=sin\beta cos(-\frac{\pi }{3} )-(cos\beta ) \cdot sin(-\frac{\pi }{3} )=sin\beta \cdot \frac{1}{2} -cos\beta \cdot (-\frac{\sqrt{3} }{2} )=\frac{sin\beta }{2} -(-\frac{1}{3} ) \cdot (-\frac{\sqrt{3} }{2} )=(*)[/tex]
Obliczymy teraz [tex]sin\beta[/tex]
[tex]sin^2\beta +cos^2\beta =1[/tex]
[tex]sin^2\beta =1-(-\frac{1}{3})^2=1-\frac{1}{9} =\frac{8}{9}[/tex]
[tex]sin\beta =\frac{\sqrt{8} }{3} =\frac{2\sqrt{2} }{3}[/tex] ( odrzucamy)
lub [tex]sin\beta =-\frac{2\sqrt{2} }{3}[/tex]
[tex](*)=\frac{\frac{-2\sqrt{2} }{3} }{2} -\frac{\sqrt{3} }{6} =\frac{-2\sqrt{2} }{6} -\frac{\sqrt{3} }{6} =\frac{-2\sqrt{2}-\sqrt{3} }{6}[/tex]
