Алгебра: с 3-м заданием. Буду очень благодарен

с 3-м заданием. Буду очень благодарен

Ответы

Hist18 15.05.2021 21:37

$1)\frac{Cos\frac{\pi }{6}-\sqrt{3}*tg60^{0}}{Sin\frac{\pi }{6}+Cos60^{0}}=\frac{\frac{\sqrt{3}}{2}-\sqrt{3}*\sqrt{3}}{\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3} }{2}-3=\boxed{\frac{\sqrt{3}-6 }{2}}\\\\2)\frac{\sqrt{3}*Ctg30^{0}+\sqrt{2}*Sin\frac{\pi }{4}}{2tg45^{0}-Cos0^{0}}=\frac{\sqrt{3}*\sqrt{3}+\sqrt{2}*\frac{1}{\sqrt{2}}}{2*1-1}=3+1=\boxed4$

$3)\frac{tg30^{0}+Cos\frac{\pi }{6}}{Sin\frac{\pi }{2}-4Ctg45^{0}}=\frac{\frac{\sqrt{3} }{3}+\frac{\sqrt{3} }{2}}{1-4*1}=\frac{\frac{5\sqrt{3} }{6}}{-3} =\boxed{-\frac{5\sqrt{3}}{18}}$

$4)\frac{\sqrt{2}*Sin45^{0}+\sqrt{2}*Cos\frac{\pi }{4}}{5*tg\frac{\pi }{4}-4*Ctg\frac{\pi }{4}} =\frac{\sqrt{2}*\frac{1}{\sqrt{2}} +\sqrt{2}*\frac{1}{\sqrt{2}}}{5*1-4*1}=\frac{1+1}{1}=\boxed2$

$5) 6Cos40^{0} -8Cos^{3} 40^{0}=2Cos40^{0}(3-4Cos^{2}40^{0})=\\\\=2Cos40^{0}(3-4*\frac{1+Cos80^{0} }{2})=2Cos40^{0}[3-2(1+Cos80^{0})] =\\\\=2Cos40^{0}(3-2-2Cos80^{0})=2Cos40^{0}(1-2Cos80^{0})=\\\\=2Cos40^{0}-4Cos40^{0}*Cos80^{0} }=2Cos40^{0}-4*\frac{Cos(40^{0} +80^{0} )+Cos(40^{0} -80^{0})}{2} =\\\\=2Cos40^{0}-2(Cos120^{0} +Cos40^{0} )=2Cos40^{0}-2Cos120^{0} -2Cos40^{0}=\\\\=-2Cos120^{0}=-2*(-\frac{1}{2}=\boxed1$

$6)\frac{4Sin25^{0}*Sin65^{0}}{Cos40^{0} }=\frac{4Sin25^{0}*Sin(90^{0}-25^{0}) }{Cos40^{0} }=\frac{4Sin25^{0}*Cos25^{0}}{Cos40^{0} } =\frac{2Sin50^{0} }{Cos40^{0} }=\\\\=\frac{2Sin(90^{0}-40^{0})}{Cos40^{0} }=\frac{2Cos40^{0} }{Cos40^{0}}=\boxed2$