Question

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ЕСЛИ БУДЕТ ВСЕ ПРАВИЛЬНО ОТМЕЧУ, КАК ЛУЧШИЙ ОТВЕТ (НУЖНО РЕШИТЬ ДЛЯ ПРОВЕРКИ)

Answer 1

$1)Sin\frac{7\pi }{3}=Sin(2\pi+\frac{\pi }{3})=Sin\frac{\pi }{3}=\frac{\sqrt{3} }{2}\\\\Cos(-\frac{5\pi }{4})=Cos(\pi+\frac{\pi }{4})=-Cos\frac{\pi }{4}=-\frac{\sqrt{2}}{2}\\\\tg(-\frac{13\pi }{6})=-tg(2\pi+\frac{\pi }{6} )=-tg\frac{\pi }{6}=-\frac{\sqrt{3}}{3}\\\\Ctg13,5\pi=Ctg(13\pi+\frac{\pi }{2})=Ctg\frac{\pi }{2}=0$

$2Sin870^{0}+\sqrt{12}Cos570^{0}-tg^{2}60^{0}=\\\\=2Sin(720^{0}+150^{0})+2\sqrt{3}Cos(360^{0}+210^{0})-(\sqrt{3})^{2}= \\\\=2Sin150^{0}+2\sqrt{3} Cos210^{0}-3=2Sin(180^{0}-30^{0})+2\sqrt{3}Cos(180^{0}+30^{0})-3=\\\\=2Sin30^{0}-2\sqrt{3}Cos30^{0}-3=2*\frac{1}{2} -2\sqrt{3}*\frac{\sqrt{3}}{2}-3=1-3-3=-5$

$2)Ctgt*Sin(-t)+Cos(2\pi-t)=\frac{Cost}{Sint}*(-Sint)+Cost=-Cost+Cost=0\\\\\\3)\frac{\pi }{2}<t<\pi\Rightarrow Cost<0\\\\Cost=-\sqrt{1-Sin^{2} t} =-\sqrt{1-(\frac{4}{5})^{2}}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\\\\tg=\frac{Sint}{Cost}=\frac{\frac{4}{5} }{-\frac{3}{5}} =-\frac{4*5}{5*3}=-\frac{4}{3}=-1\frac{1}{3} \\\\Ctgt=\frac{1}{tgt}=\frac{1}{-\frac{4}{3}} =-\frac{3}{4}=-0,75$

$4)\frac{Ctgt}{tgt+Ctgt}=\frac{\frac{Cost}{Sint} }{\frac{Sint}{Cost}+\frac{Cost}{Sint}}=\frac{\frac{Cost}{Sint} }{\frac{Sin^{2}t+Cos^{2}t}{Sint Cost}}=\frac{\frac{Cost}{Sint} }{\frac{1}{Sint Cost}}=\frac{Cost*SintCost}{Sint}=Cos^{2}t\\\\Cos^{2} t=Cos^{2}t$