Odpowiedź:
[tex]4\sqrt{3}sin\alpha cos\alpha =3\sqrt{2}-2\sqrt{3}[/tex]
Szczegółowe wyjaśnienie:
[tex](sin\alpha +cos\alpha )^{2}=\frac{\sqrt{6} }{2}\\sin^{2}\alpha +2sin\alpha cos\alpha +cos^{2}\alpha = \frac{\sqrt{6} }{2}\\1+2sin\alpha cos\alpha =\frac{\sqrt{6} }{2}\\2sin\alpha cos\alpha =\frac{\sqrt{6} }{2}-1=\frac{\sqrt{6}-2 }{2} \\[/tex]
Zatem:
[tex]4\sqrt{3}sin\alpha cos\alpha =4\sqrt{3}=2\sqrt{3} *2sin\alpha cos\alpha =2\sqrt{3}*\frac{\sqrt{6}-2 }{2}=\sqrt{3}(\sqrt{6}-2)=\sqrt{18} -2\sqrt{3}=3\sqrt{2} -2\sqrt{3}[/tex]
