Question

помогите пожалуйста решить. даю 11 баллов!!!

Answer 1

Ответ:

4).

$= {sin}^{2} x$

5).

= - 1,5

Пошаговое объяснение:

4).

${ctg}^{2} x \times {sin}^{2} x - cos2x = {( \frac{cosx}{sinx})}^{2} \times {sin}^{2} x - cos2x = \frac{ {cos}^{2}x}{ {sin}^{2}x} \times {sin}^{2} x - ( {cos}^{2} x - {sin}^{2} x) = {cos}^{2} x - {cos}^{2} x + {sin}^{2} x = {sin}^{2} x$

5).

$3ctg {60}^{0}\times (sin {310}^{0} cos {70}^{0} - cos {310}^{0} sin {70}^{0} ) = 3ctg {60}^{0} \times sin( {310}^{0} - {70}^{0}) = 3ctg {60}^{0} \times sin {240}^{0} = 3ctg {60}^{0} \times sin( {270}^{0}-{30}^{0}) = 3ctg {60}^{0} \times ( - cos {30}^{0}) = 3 \times \frac{ \sqrt{3} }{3} \times ( - \frac{ \sqrt{3} }{2} ) = - 1.5$