Предмет: Алгебра, автор: kmezenceva931

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Приложения:

slava13213: Под б : верхняя дробь cot2(α)(sin(α)+1)(sin(α)−1) нижняя дробь (cos(α)+1)(cos(α)−1)

Ответы

Автор ответа: Universalka

$\displaystyle\bf\\1)\\\\Cos\frac{2\alpha }{3} \cdot Sin\frac{2\alpha }{3} \cdot Cos\frac{4\alpha }{3} =\frac{1}{2} \cdot\Bigg(2Cos\frac{2\alpha }{3} \cdot Sin\frac{2\alpha }{3} \Bigg)\cdot Cos\frac{4\alpha }{3} =\\\\\\=\frac{1}{2} \cdot Sin\frac{4\alpha }{3} \cdot Cos\frac{4\alpha }{3} =\frac{1}{2} \cdot \frac{1}{2} \cdot \Bigg(2Sin\frac{4\alpha }{3} \cdot Cos\frac{4\alpha }{3} \Bigg)=\\\\\\=\frac{1}{4} Sin\frac{8\alpha }{3} =0,25Sin\frac{8\alpha }{3}$

$\displaystyle\bf\\2)\\\\\frac{Sin^{2} 2\alpha -4Cos^{2} \alpha }{Sin^{2} 2\alpha +4Cos^{2} \alpha -4} =\frac{4Sin^{2}\alpha Cos^{2} \alpha -4Cos^{2} \alpha }{4Sin^{2}\alpha Cos^{2}\alpha +4Cos^{2} \alpha -4} =\frac{4 Cos^{2} \alpha(Sin^{2} \alpha -1)}{4(Sin^{2}\alpha Cos^{2} \alpha +Cos^{2} \alpha -1)} =\\\\\\=\frac{Cos^{2} \alpha \cdot(-Cos^{2}\alpha ) }{Sin^{2}\alpha Cos^{2} \alpha -Sin^{2} \alpha } =\frac{-Cos^{4}\alpha }{Sin^{2} \alpha\cdot (Cos^{2} \alpha -1)} =$