Question

Помогите пожалуйста срочно

Answer 1

$\displaystyle\bf\\5)\\\\\frac{3\pi }{2} < \alpha < 2\pi \ \ \ \Rightarrow \ \ \ Sin\alpha < 0 \\\\\pi < \beta < \frac{3\pi }{2} \ \ \ \Rightarrow \ \ \ Cos\beta < 0\\\\\\Sin\alpha =-\sqrt{1-Cos^{2}\alpha } =-\sqrt{1-\Big(\frac{3}{5} \Big)^{2} } =-\sqrt{1-\frac{9}{25} } =\\\\\\=-\sqrt{\frac{16}{25} } =-\frac{4}{5} \\\\\\Cos\beta =-\sqrt{1-Sin^{2}\beta } =-\sqrt{1-\Big(-\frac{8}{17} \Big)^{2} } =-\sqrt{1-\frac{64}{289} } =\\\\\\=-\sqrt{\frac{225}{289} } =-\frac{15}{17}$

$\displaystyle\bf\\Sin\Big(\alpha -\beta \Big)=Sin\alpha Cos\beta -Cos\alpha Sin\beta =-\frac{4}{5} \cdot\Big(-\frac{15}{17} \Big)-\frac{3}{5} \cdot\Big(-\frac{8}{17}\Big)=\\\\\\=\frac{60}{85} +\frac{24}{85} =\frac{84}{85} \\\\\\6)\\\\\frac{\Big[Sin\Big(\pi -3\alpha \Big)-Cos\Big(\dfrac{3\pi }{2}+\alpha \Big)\Big] \Big[Sin\Big(\dfrac{\pi }{2}+3\alpha \Big) +Cos\Big(\pi +\alpha \Big)\Big]}{1+Cos\Big(\pi -2\alpha \Big)} =$

$\displaystyle\bf\\=\frac{\Big(Sin3\alpha -Sin\alpha \Big)\cdot\Big(Cos3\alpha -Cos\alpha\Big) }{1-Cos2\alpha } =\\\\\\=\frac{2Sin\dfrac{3\alpha -\alpha }{2}Cos\dfrac{3\alpha +\alpha }{2} \cdot\Big(-2Sin\dfrac{3\alpha -\alpha }{2} \Big)\cdot Sin\dfrac{3\alpha +\alpha }{2} }{1-Cos2\alpha } =\\\\\\=-\frac{4Sin\alpha Cos2\alpha Sin\alpha Sin2\alpha }{2Sin^{2} \alpha } =-2Sin2\alpha Cos2\alpha =-Sin4\alpha \\\\\\-Sin4\alpha =-Sin4\alpha$

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