Question

√120*сos(-4π/3)*sin(-π/4)

Answer 1

Ответ:

$\sqrt{120}\cdot cos(-\dfrac{4\pi }{3})\cdot sin(-\dfrac{\pi}{4})=\sqrt{120}\cdot cos(\pi +\dfrac{\pi}{3})\cdot (-sin\dfrac{\pi}{4})=\\\\=-2\sqrt{30}\cdot (-cos\dfrac{\pi}{3})\cdot \dfrac{\sqrt2}{2}=2\sqrt{30}\cdot \dfrac{1}{2}\cdot \dfrac{\sqrt2}{2}=\dfrac{\sqrt{60}}{2}=\dfrac{2\sqrt{15}}{2}=\sqrt{15}$

Answer 2

Ответ:

√120*cos(-4π/3)*sin(-π/4)=√120*(-½)*(-√2/2)=2√30*½*√2/2=√30*√2/2=√60/2=(2√15)/2=√15.