.........задания на фото

Решите задачу:

$1)8\sqrt{3}Cos^{2} \frac{7\pi }{12}-8\sqrt{3}Sin^{2}\frac{7\pi }{12}=8\sqrt{3}(Cos^{2}\frac{7\pi }{12} -Sin^{2}\frac{7\pi }{12})=8\sqrt{3}Cos(2*\frac{7\pi }{12})=8\sqrt{3}Cos\frac{7\pi }{6}=8\sqrt{3}Cos(\pi+\frac{\pi }{6})=8\sqrt{3}*(-Cos\frac{\pi }{6})=-8\sqrt{3}*\frac{\sqrt{3} }{2}=-12$

$2)8\sqrt{3}-8\sqrt{12}Sin^{2} \frac{5\pi }{12}=8\sqrt{3}-8\sqrt{4*3}Sin^{2}\frac{5\pi }{12}=8\sqrt{3}-16\sqrt{3}Sin^{2}\frac{5\pi }{12} =8\sqrt{3}(1-2Sin^{2}\frac{5\pi }{12})=8\sqrt{3}*Cos(2*\frac{5\pi }{12}) }=8\sqrt{3}Cos\frac{5\pi }{6} =8\sqrt{3} Cos(\pi- \frac{\pi }{6})=8\sqrt{3}*(-Cos\frac{\pi }{6})=-8\sqrt{3}*\frac{\sqrt{3} }{2}=-12$

$3)-95Cos2\alpha=-95*(2Cos^{2}\alpha-1)=-95(2*(-0,6)^{2}-1)=-95*(0,72-1)=-95*(-0,28)=2,66$

.........задания ** фото