Question

Алгебра, 10
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Answer 1

$1)\sqrt{3} \cdot \Big(Cos\frac{11\pi }{12}-Cos\frac{5\pi }{12} \Big)=\sqrt{3}\cdot \Big(-2Sin\frac{\frac{11\pi }{12}+\frac{5\pi }{12}}{2}\cdot Sin\frac{\frac{11\pi }{12}-\frac{5\pi }{12}}{2}\Big)=\\\\=\sqrt{3}\cdot\Big(-2Sin\frac{2\pi }{3} \cdot Sin\frac{\pi }{4}\Big)=-2\sqrt{3}Sin(\pi-\frac{\pi }{3} )\cdot \frac{\sqrt{2} }{2}=-\sqrt{6}Sin\frac{\pi }{3} =\\\\=-\sqrt{6}\cdot \frac{\sqrt{3} }{2}=\boxed{-\frac{3\sqrt{2} }{2} }$

$2)Sin\frac{7\pi }{12} Cos\frac{\pi }{12}-Sin\frac{\pi }{12}Cos\frac{7\pi }{12}=Sin(\frac{7\pi }{12} -\frac{\pi }{12})=Sin\frac{6\pi }{12}=Sin\frac{\pi }{2}=\boxed1$

$3)\frac{3\pi }{2}<\alpha<2\pi \ \Rightarrow Cos\alpha>0\\\\0<\beta <\frac{\pi }{2} \ \Rightarrow \ Cos\beta>0\\\\Cos\beta=\sqrt{1-Sin^{2}\beta}=\sqrt{1-(\frac{8}{17})^{2}}=\sqrt{1-\frac{64}{289} } =\sqrt{\frac{225}{289} }=\frac{15}{17}\\\\Cos\alpha=\sqrt{1-Sin^{2}\beta}=\sqrt{1-(-\frac{3}{5})^{2}}=\sqrt{1-\frac{9}{25} } =\sqrt{\frac{16}{25} }=\frac{4}{5} \\\\85\cdot Cos(\alpha +\beta)=85\cdot (Cos\alpha Cos\beta-Sin\alpha Sin\beta)=$

$=85\cdot\Big(\frac{4}{5} \cdot \frac{15}{17}+(-\frac{3}{5})\cdot \frac{8}{17}\Big)=85\cdot\Big(\frac{60}{85}-\frac{24}{85}\Big)=85\cdot \frac{36}{85}=\boxed{36}$