Question

cos120°-ctg 60°+sin 90°cos 135°

Answer 1

Ответ:

$cos120^\circ -ctg60^\circ +sin90^\circ \cdot cos135^\circ =cos(180^\circ -60^\circ )-\dfrac{\sqrt3}{3}+1\cdot cos(90^\circ +45^\circ )=\\\\=-cos60^\circ -\dfrac{\sqrt3}{3}-sin45^\circ =-\dfrac{1}{2}-\dfrac{\sqrt3}{3}-\dfrac{\sqrt2}{2}=-\dfrac{3+2\sqrt3+3\sqrt2}{6}$