Question

cosπ/6-√3tg60°/sinπ/6+cos60°​

Answer 1

$1)\frac{Cos\frac{\pi }{6}-\sqrt{3}tg60^{0}}{Sin\frac{\pi }{6}+Cos60^{0}}=\frac{\frac{\sqrt{3} }{2}-\sqrt{3}*\sqrt{3}}{\frac{1}{2}+\frac{1}{2}}=\frac{\frac{\sqrt{3}}{2}-3 }{\frac{1}{4} }=\boxed{2(\sqrt{3}-6)}\\\\\\2)\frac{tg30^{0}+Cos\frac{\pi }{6}}{Sin\frac{\pi }{2}-4Ctg45^{0}}=\frac{\frac{\sqrt{3} }{3}+\frac{\sqrt{3}}{2}}{1-4*\frac{\sqrt{2} }{2}}=\frac{\frac{5\sqrt{3}}{6} }{1-2\sqrt{2}} =\boxed{\frac{5\sqrt{3} }{6(1-2\sqrt{2})}}$

При решении третьего задания применим формулу :

Cos3x = 4Cos³x - 3Cosx

$6Cos40^{0}-8Cos^{3}40^{0}=-2(4Cos^{3}40^{0}-3Cos40^{0})=\\\\=-2Cos(3*40^{0})=-2Cos120^{0}=-2*(-\frac{1}{2})=\boxed1$