Question

Помогите с алгеброй 2 задания - 100 баллов. Дать полный ответ

Answer 1

$\displaystyle\bf\\7)\\\\\frac{3\pi }{2} < \alpha < 2\pi \ \ \ \Rightarrow \ \ \ Cos\alpha > 0\\\\\\Cos\alpha =\sqrt{1-Sin^{2}\alpha } =\sqrt{1-\Big(-\frac{4}{5}\Big)^{2} } =\sqrt{1-\frac{16}{25} } =\sqrt{\frac{9}{25} } =\frac{3}{5} =0,6\\\\\\8)\\\\\frac{Sin(30^\circ+\alpha )-Cos(60^\circ+\alpha )}{Sin(30^\circ+\alpha )+Cos(60^\circ+\alpha )} =\frac{Cos\Big[90^\circ-(30^\circ+\alpha )\Big]-Cos(60^\circ+\alpha )}{Cos\Big[90^\circ-(30^\circ+\alpha )\Big]+Cos(60^\circ+\alpha) } =$

$\displaystyle\bf\\=\frac{Cos(60^\circ-\alpha )-Cos(60^\circ+\alpha )}{Cos(60^\circ-\alpha )+Cos(60^\circ+\alpha )} =\\\\\\=\frac{-2Sin\dfrac{60^\circ-\alpha -60^\circ-\alpha }{2} Sin\dfrac{60^\circ-\alpha +60^\circ+\alpha }{2} }{2Cos\dfrac{60^\circ-\alpha -60^\circ-\alpha }{2}Cos\dfrac{60^\circ-\alpha +60^\circ+\alpha }{2} } =\\\\\\=\frac{-2Sin(-2\alpha )\cdot Sin120^\circ}{2Cos(-2\alpha )\cdot Cos120^\circ} =$

$\displaystyle\bf\\=\frac{Sin2\alpha \cdot\frac{\sqrt{3} }{2} }{Cos2\alpha \cdot(-\frac{1}{2}) }=-\sqrt{3} tg2\alpha$