Question

Помогите , срочно надо .

Answer 1

Ответ:

$1)\ \ 2\, sin120^\circ +4\, cos150^\circ -2\, tg135^\circ =2\, cos30^\circ -4cos30^\circ +2ctg45^\circ =\\\\=2\cdot \dfrac{\sqrt3}{2}-4\cdot \dfrac{\sqrt3}{2}+2\cdot 1=\sqrt3-2\sqrt3+2=-\sqrt3+2\\\\\\2)\ \ cos120^\circ -8\, sin^2150^\circ +3\, cos90^\circ \cdot cos162^\circ =-sin30^\circ -8\cdot sin^230^\circ +3\cdot 0=\\\\=-\dfrac{1}{2}-8\cdot \dfrac{1}{4}+0=-\dfrac{1}{2}-2=-2\dfrac{1}{2}$

$3)\ \ cos180^\circ \cdot (sin135^\circ \cdot tg60^\circ -cos135^\circ )^2=-1\cdot (cos45^\circ \cdot \sqrt3+sin45^\circ)^2=\\\\=-\Big(\dfrac{\sqrt2}{2}\cdot \sqrt3+\dfrac{\sqrt2}{2}\Big)^2=-\Big(\dfrac{\sqrt3}{\sqrt2}+\dfrac{1}{\sqrt2}\Big)^2=-\Big(\dfrac{\sqrt3+1}{\sqrt2}\Big)^2=-\dfrac{4+2\sqrt3}{2}=\\\\=-2-\sqrt3$

Answer 2

Ответ:

Объяснение:

1)  2sin 120° + 4cos 150° - 2tg 135° =

 = 2 sin (90° + 30°) + 4cos (90° + 60°) - 2tg (90° + 45°) =

= 2 cos 30° - 4 sin 60° +2 ctg 45° = 2 · 0.5√3 - 4 · 0.5√3 + 2 · 1 =

= √3 - 2√3 + 2 =

= 2 - √3

2) cos 120° - 8 sin²150° + 3cos 90° · cos 162° =

  = cos (90° + 30°) - 8 sin²(90° + 60°) + 3 · 0 · cos 162° =

  = - sin 30° - 8 · cos² 60° + 0 =

  = -0.5 - 8 · (0.5)² =

  = - 2.5

3) cos 180° · (sin 135° · tg 60° - cos 135°)² =

  = (-1) · (sin (90° + 45°) · √3 - cos (90° + 45°))² =

  =  - (cos 45°  · √3 + sin 45°)² =

  = - (0.5 √2 · √3 + 0.5 √2)²=

  = -( 0.5√2 · (√3 + 1))² =

  = - ( 0.25 · 2 · (3 + 2√3 + 1) =

  = - (0.5 · (4 + 2√3) =

  = - (2 + √3)

  = - 2 - √3